`
LULU
2007
High Speed Calculations
1
Rajnish Kumar, born in 1971, is a Mechanical Engineer working with the Indian
Railways. He calls himself a number maniac, but never got cent percent in Mathematics.
He loves the subject and after years of exciting adventures with figures and numbers
treads upon the path of writing on Mathematics.
High Speed Calculations
© Rajnish Kumar, 2006
All knowledge imparted through this book is in public domain.
2
3
High Speed Calculations
CONTENTS
What this book can do for you? .......................................................................................... 4
Scheme of the book ............................................................................................................. 8
CHAPTER 1
Quick Addition Techniques .................................................................. 10
CHAPTER 2
Techniques For Memorising Multiplication Tables.............................. 17
CHAPTER 3
Speeding up Basic Multiplication ......................................................... 21
CHAPTER 4
Multiplication of two digits................................................................... 26
CHAPTER 5
Higher Order Multiplication.................................................................. 32
CHAPTER 6
Simple Division..................................................................................... 36
CHAPTER 7
More Division ....................................................................................... 42
CHAPTER 8
Quick Squaring...................................................................................... 52
CHAPTER 9
Square Roots ......................................................................................... 59
CHAPTER 10
Quick Cubes .......................................................................................... 63
CHAPTER 11
Cube Roots ............................................................................................ 67
CHAPTER 12
Recurring Decimals............................................................................... 72
CHAPTER 13
Divisibility Rule For Any Number ....................................................... 75
CHAPTER 14
Quick Averaging ................................................................................... 81
CHAPTER 15
Application of Techniques to Algebraic equations ............................... 84
Solutions to Exercises in Chapters .................................................................................... 89
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High Speed Calculations
What this book can do for you?
The single biggest fear for all students of mathematics is the cumbersome calculations
involving even medium sized numbers. The reason being two fold
1. Lack of time.
2. No intrinsic way for verification of result.
The problem is aggravated more during competitive examinations where speed and
accuracy are the essence, and even a slight error can fetch negative marks. This simply is
overwhelming because of which lot of escapist tendencies and remorse occurs in
potential candidates. This book has been compiled after extensive study of various quick
calculation techniques picked up from Prof Tratchenberg’s system for speed Mathematics
and Vedic Mathematics, and also from tidbits across a spectrum of books. It has been
kept very simple and only the very relevant and “truly” QUICK techniques have been
mentioned. Most of the exercises have been fully solved highlighting all important steps
and potential areas of error. The techniques increase speed and also provide an intrinsic
verification of result.
The wonders that you can do with your calculations are
•
Add up 11342+67545+89876+56428+78625+23148 in just 30 seconds!
•
Without jotting down a single digit, mentally multiply 54x67, 89x12, 87x14 etc.
•
In single line multiply 89x98, 95x86, 88x99 etc.
•
You can even multiply 544x567=308448 in a single step
•
357x732= 2 1 4 4 7 0 3 1 1 4 ⇒ 261324
•
62÷9 without dividing, don’t believe this, continue reading!
•
10113÷88 with only addition and plain single digit multiplication
Great so easy!! Just one line
88 ) 101 / 13
12
1/2
/ 48
______
114 / 81
•
More surprising 98613÷63 can be found with just division by 6, like this
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High Speed Calculations
3
9
6
8
6
3
1
1
5
/
5
5
3
3
6
5
/
18
Quotient = 1565, Remainder = 18
•
Even more interesting 25453÷8829 in just one step
8 8 2 9
1 1 7 1
1 2 -3 1
2
/ 5
2
4
4
5
-6
3
2
2
/ 7
7
9
5
Quotient = 2, Remainder = 7795
•
You can find square of 97 in just a blink
972 = 97-3/09 = 9409
10092 = 1009+9/081 = 1018081, unbelievable but possible
•
If this does not appeal can you find the square of 7693 in one line, without several
steps involving multiplication and addition.
76932 = …7693 ⇒ 49 84
•
162 150 117 54
9 ⇒ 59182249
The only way to find a square root is using a calculator which is not actually
allowed upto Intermediate examination and in competition. But there is two-line
method to find square root of any number.
Square root of 20457529 is 4523 by this two-line method
2
0
:
4
4
8
4
5
4
5
:
7
5
4
2
3
3
:
:
2
1
0
9
0
0
0
High Speed Calculations
•
6
123 can be calculated without multiplying twice
123= 1 2+4 4+8 8 = 1728 in just this much space
•
Also 9973 = 991/027/-027 = 991026973, can you imagine the reduction in
calculation time with these steps?
•
Look at the number and find the cube root
For 262,144 the cube root is 64 without a single calculation
And for 571,787 cube root is 83, believe this book and you will be a calcu-magician.
•
With a little more effort the cube root of 8 to 9 digit numbers can be found, for
which there is no standard method in conventional arithmetic. For example,
Cube root of 178,453,547 is 563 in just two steps.
a=3 , c=5 , 178453547-27= 178453520,
3x32b end in 6 thus b=2, so cube root is 563.
Confused! You shall not be once all chapters are studied and problems practiced.
•
Did you know
1/19 = . 0 5 2 6 3 1 5 7 8 9 4 7 3 6 8 4 2 1
can be calculated in just this single step
and also
1/29= ………19 16 15 5 21 7 12 4 11 23 27 9 3 1
=. 0 3 4 4 8 2 7 5 8 6 2 0 6 8 9 6 5 5 1 7 2 4 1 3 7 9 3 1
Amazing, that’s the power of this book.
There are a few more things to know. All of us have heard of divisibility rules for
2,3,4,5,6,8,9,11 etc. These are generally taught in all schools. But there is actually a rule
for any number whatsoever and its really easy, much more than you can imagine.
To find whether 10112124 is divisible by 33 all you have to do is
1011212+40=1011252⇒101125+20=101145⇒10114+50=10164⇒1016+40⇒
1056⇒105+60=165⇒16+50=66 : and you have the answer, with only simple addition
and multiplication to do and no cumbersome division till the end.
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High Speed Calculations
Using the clichéd statement, you have to learn to believe. There is enough potential in
you to become a mathe-magician rather a calcu-magician. All that is needed is
wholehearted attempt without the slightest doubt and cynicism on mastering the
techniques explained.
The author made a rough estimate that in the arithmetic portion of GRE, GMAT etc.,
there is possibility of arriving at right solution by intuition in 10% of the cases, and 90%
of the time is WASTED in calculations. The book can help reduce this time by as much
as 50% for a person with normal arithmetical skills and a person with genius can actually
be near to paperless in calculations, of course jotting down the answer is excluded. This
is not a false claim. Geniuses like Shakuntla Devi from India have only mastered such
techniques to become calculation wizards.
Happy calculating.
WARNING: Do not attempt to reduce time further as these are the shortest possible
ways to calculate, any attempt on that line can be hazardous and self-defeating.
Rajnish Kumar
rajnishkumar1971@yahoo.co.in
High Speed Calculations
8
Scheme of the book
The book contains chapters on following aspects
1. Quick Addition techniques
2. Techniques for memorizing Multiplication tables.
3. Speeding up basic Multiplication.
4. Multiplication of two digits by two digits.
5. Higher order Multiplication Techniques
6. Simple Division
7. Higher order Division
8. Checking the Division
9. Quick Squaring of numbers
10. Quick Square roots
11. Quick Cubing of numbers
12. Quick Cube roots
13. Recurring Decimals
14. Rules for Divisibility for any number - Positive Osculation Method
15. Rules for Divisibility for any number - Negative Osculation Method
16. Quick Averaging of numbers
17. Application of techniques to Algebraic equations.
Quick mathematics. It sounds a bit unusual but it is not. They have been very intelligent
and innovative teachers in mathematics who have developed very nice ways of teaching
unusual ways of calculations. It is often bewildering for the mother to teach the
multiplication tables which later on help in calculations. They can be a special way to
train the fresh mind on calculations, which may later on retain the interest and also ease
during the final stages of schooling. Most of the competitive examinations do not allow
the calculator. Whether it is the famous GRE,GMAT,CAT or any of the banking
examination, the students who are slightly poor at calculations become part of the great
depression circle. The numbers scare them and the scare prevents them from improving
their skills. The scheme in the book may not be replacing any of the conventional
techniques but it can certainly add on and complement whatever gaps are found in the
conventional technique. It is important not to be overenthusiastic about what is learnt
their but adapt the skills in a manner that for a particular situation the technique can be
applied and tremendous saving in time is achieved. Nothing can replace the older proven
calculation techniques, as they are universal by nature. The techniques depicted in the
book are for particular set of number and are really effective in saving time.
Besides the practical utility of the techniques, it will also be very interesting for those
who love numbers and their relationships. Each technique is actually a very simple
algebraic proof. The proof is not dealt in this book because it is of mere academic
interest and does not really help in easing calculations.
High Speed Calculations
9
The only difficulty with these techniques is that they are too many and often it may be
confusing for the student to remember all of them. But there is a way out. The exercises
at the end of the chapter should be done at least three to four times before leaving the
chapter. This way the technique is ingrained and will like a reflex action of a car driver
automatically pressing the correct pedal.
The difficulties arising out of quick calculations are of different nature. The first problem
is with the addition of subsequent numbers that sometimes leads to error if there is any
break in the concentration. So it is most important to be particularly attentive during the
calculation and nothing else should be thought of. Quick calculation does not mean
impatience or haste, it should be done very systematically and strictly as per the
guidance. This is the shortest possible way to calculate, so if the student again tries to
shorten the procedure he would land at making silly mistakes and of course blame the
poor book for the misdemeanor.
The second problem is with the addition and subtraction involved. During the course of
learning these techniques it is very important to have sound practice of addition and
subtraction of two numbers. There should be absolutely no waste of time in adding or
subtracting two numbers. This aspect should be thoroughly practiced to perfection.
Hopefully the book will serve its purpose.
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High Speed Calculations
CHAPTER 1
Quick Addition Techniques
The basic thing for practice to get accurate calculations is addition. It is the easiest but the
most error producing aspect of quick calculations. The addition of two single digit
numbers should be just a REFLEX Action. Following is a method to develop this reflex.
Place all numbers 1 to 9 in a grid.
5
2
9
4
1
6
8
3
7
1
2
3
4
5
6
7
8
9
Now quickly without wasting any second fill in the columns adding the number in the
row and column. (Notice the numbers in the column are not in sequence). Anyone who
is not too good at calculations will make some mistakes. Not to worry, anyone who is not
good at calculations will try to be careful and try to relate the calculation to the previous
one. So do this again until there is perfection. The ideal time for filling up the grid 150
seconds without a single mistake. Practice this to perfection and then when you see two
numbers the addition will be reflex action.
One more similar exercise for perfecting addition is to mark the wrong answers in the
grid below within 150 seconds using a stopwatch.
4
2
9
3
5
6
1
8
7
5 2 9 4 1 6 8 3 7
9 7 11 8 5 10 12 3 9
7 4 11 6 3 8 5 10 9
14 11 18 13 10 15 17 12 16
8 5 12 7 4 9 11 6 10
10 7 14 9 6 11 13 8 11
10 8 15 10 7 11 14 9 13
6 3 10 5 2 7 9 4 8
13 10 17 12 9 14 16 11 15
12 9 16 11 8 13 15 10 14
There is all the possibility that you will loose patience half way, but that's the whole
game. Do not loose patience even if you eyes get sort of rolled over. Once this grid is
11
High Speed Calculations
over, you can ask your friend to make a grid for you incorporating some mistakes so that
enough practice is done and by being able to recognize errors your capability to make
error free additions will improve.
EXERCISE 1.1
Add up these figures mentally and jot down on a paper
Time 10 seconds
3+4 5+7 7+5 7+8 3+6 9+8 1+5 9+8 8+8 6+4
How many mistakes did you make?
EXERCISE 1.2
Copy and fill up the grid.
Time 30 seconds
7
2
6
4
3
4
2
9
3
5
6
1
8
7
How many mistakes did you make? In case you could not complete the grid in 30
seconds, do it again and again till you reach the 30 seconds figure.
Now we shall proceed to some techniques for additions of higher order, where this
REFLEX addition will be the only thing required. It is the tendency of most to be very
sure of their common mistakes. For example, if you are asked to calculate 6 x 7, the
answer would be 42 but while making a continuous calculation it may be done as 61 or
62, thus rendering the whole effort useless.
Adding up numbers like the following example is quite easy but there is enough scope for
mistakes as 6+7+7+2 all digits of units place add up to 22, then 2 is carried over to ten's
place and so on the addition is done in the conventional way. The grocery shopkeeper can
often make mistakes in case he is without calculator thus putting you at a loss most of the
times!
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High Speed Calculations
586
657
897
542
456
3138
Considering the same example we shall learn the Trachtenberg's Addition System.
THE RULE IS
Do not add more after Eleven.
While adding up numbers as soon as the total gets more than 11, STOP. Reduce the
figure by 11, tick mark and proceed further. Do not get confused, just pay attention
carefully.
FIRST COLUMN UNIT'S PLACE
586
657 tick
897
542 tick
456
add 6+7=13 drop 11
2 is left
now add 2+7=9
add 9+2=11 drop 11, 0 is left
add 0+6=6
we have 2 ticks and 6 as sum of
first column
SECOND COLUMN TEN'S PLACE
586
657 tick
897 tick
542
456
add 8+5=13 drop 11
2 is left
now add 2+9=11 drop 11, 0 is left
add 0+4=4
add 4+5=9
we have 2 ticks and 9 as sum of
second column
THIRD COLUMN HUNDRED'S PLACE
586
657 tick
897
542 tick
456
add 5+6=11 drop 11
0 is left
now add 0+8=8
add 8+5=13 drop 11, 2 is left
add 2+4=6
we have 2 ticks and 6 as sum of
third column
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High Speed Calculations
It is expected that the reader has understood what has been done upto this point. We have
simply isolated each column and added separately.
•
•
•
•
Whenever total is more than 11
Subtract 11 from the result ,
Place a tick
Proceed further with the result
All this can sound very complex but just be patient and you shall find this is to be quite
advantageous while making long calculations.
Running total
Ticks
5
6
8
5
4
8
5
9
4
5
6
7
7
2
6
6
2
9
2
6
2
Now prefix 0 to both the rows of running total and ticks
IMPORTANT STEP
ADD THE RUNNING TOTAL AND TICKS WITH RIGHT NEIGHBOUR IN
ROW OF TICKS. This addition is based on conventional rules
Observe closely.
Running
total
0
6
9
6
Ticks
0
2
2
2
6+2+2+
1(carry over)
=11
retain 1 carry
over 1
1
9+2+2=13
retain 3 carry
over 1
6+2+0 =8 (as
there is no
neighbour)
3
8
0+0+2+1
Total
3
The explanation may seem so long and futile. Nothing to worry, look at the table below.
If the procedure is understood then there is no need for any extra calculation and all is
done mentally with the only thing to jot down is the tick and final answer.
Let us add the numbers. You first try the conventional way. Isn't this boring and very
confusing. Now let us try the method explained above.
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High Speed Calculations
Well what can make you commit an error is subtraction of 11
•
•
simply remove the first digit.
reduce remaining by 1 e.g. 18, remove 1 and subtract 1 from 8 making it 7.
4
6
7
3
8
3
5'
9'
8'
5'
9'
6
7'
5
4
2
3
1
4
9'
8'
7'
6
8'
9'
5
4
3'
6'
7
running
total
0
9
9
0
1
9
ticks
0
2
2
3
3
3
adding up
Grand total
0+0+2+1= 9+2+2+1=
1+3+3+1=
9+2+3=14 0+3+3=6
3
14
8
3
4
4
6
8
9+3=12
2
Notice the' marks which have been used as ticks. If you examine the procedure it is all
mental. You only need to mark the tick on the number and write the running total.
For practice do the following calculations yourself. The detailed calculation is in the
solutions chapter.
EXERCISE 1.3
Add up the following using the rules explained
1.
2897+4356+7843+5434
2.
3768+5467+9087+8796
3.
34256+18976+98768+45367+13987
4.
2341+5467+9088+7009+3219
5.
14678+45325+58326+90087+76509
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High Speed Calculations
VERIFICATION OF ANSWER
While making large calculations, there is enough scope for mistakes at any stage. There
are a few very pin pointed methods of checking any calculation. But we shall discuss the
simplest one. This technique is applicable for multiplication, division, subtraction and
addition.
Let us first understand what INTEGER SUM is. It is defined as the addition of all the
numbers of the digit so that only one digit is the final result. For example,
For the number 12, add up 1 and 2 to get number 3. This is the integer sum of 12.
Let us consider a larger number like 462789,
where we start with 4+6 = 10, 10+2=12, 12+7=19, 19+8=27, 27+9= 36, and finally
3+6=9, thus 9 is the integer sum of 462789.
There are shortcuts to this,
Shortcut 1
Observe the number and whenever you find the number 9, do not consider
it for integer sum.
Shortcut 2
A group of numbers like 27, 36, 45, 81 which add up to 9, can be ignored.
The reason is when you add 9 to a number its integer sum remains the same. So instead
of wasting your effort in adding up 9 to the integer sum, it may best be ignored.
Shortcut 3
digit to add.
At each step keep adding up the number, so that there is always a single
Using this scheme let us redo the integer sum of 462789
•
•
•
•
ignore 9 and group of 7 and 2 (shortcut 1 and 2)
we are left with 4,6,8
4+6=10, 1+0=1 (shortcut 3)
now 1+8=9
Taking another example, find integer sum of 437862901
•
•
ignore 9, 6/3,7/2, 8/1
we are left with 4 and 0, hence integer sum is 4, SIMPLE !
16
High Speed Calculations
EXERCISE 1.4
Find the Integer sums of following numbers
Time 45 seconds- 1st try
30 seconds- 2nd try
1.
2.
3.
4.
5.
6.
7.
361547980
376539246
546372
900876
9994563
918273645
225162
Continuing with the topic of verification of answer. The aim of the integer check is to
first find the integer sums of all the numbers to be added and then adding up these integer
sums, to get the integer sum which should be same as integer sum of the result. The
principle is integer sum of result obtained should be same as this. The following example
using the same set of numbers would illustrate this
integer sum of
numbers
Grand
total
3
4
6
7
3
8
1
3
5'
9'
8'
5'
3
9'
6
7'
5
4
4
2
3
1
4
9'
1
8'
7'
6
8'
9'
2
5
4
3'
6'
7
7
2
9
9 is the integer
is integer
sum of this
sum of
column
344682
4
4
6
8
Here you find that the integer sum of result and of the constituents is same, hence the
result is verified. For exercise 1.3 do the same and verify your answers using this method.
17
High Speed Calculations
CHAPTER 2
Tables
Techniques For Memorising Multiplication
The most difficult aspect of calculations is memorising the multiplication tables. If not
much at least up to 20 should be on the fingertips. If one can as explained previously
develop reflexes for multiplying numbers like 13 and 8, 12 and 6 etc, then there is
tremendous scope for increasing the multiplication speed by three or four times without
using any unconventional method which shall be explained later in this book.
The first thing is to find patterns in the multiplication table. Form a grid of all the
numbers up to 10 in rows and columns just as was done for the improvement in addition.
After the normal grid is practiced, it should be jumbled up, so that random answers
emanate. The cramming up of the table should not be like that of a child who forgets the
table at the very first instance of disturbance.
1
2
3
4
5
6
7
8
9
10
1
1
2
3
4
5
6
7
8
9
10
2
2
4
6
8
10
12
14
16
18
20
3
3
6
9
12
15
18
21
24
27
30
4
4
8
12
16
20
24
28
32
36
40
5
5
10
15
20
25
30
35
40
45
50
6
6
12
18
24
30
36
42
48
54
60
7
7
14
21
28
35
42
49
56
63
70
8
8
16
24
32
40
48
56
64
72
80
9
9
18
27
36
45
54
63
72
81
90
10
10
20
30
40
50
60
70
80
90
100
Go through this, and you shall find are certain patterns, which can be remembered to
have no confusions regarding these basic one digit multiplication. If you are very quick
at small additions and multiplication then it may not be really important remember the
tables up to 20. As 13 x 8, 12 x 6, 14 x 9 can be easily be derived. But it is always an
advantage to remember these two-digit multiplications also. Complete the exercise in the
stipulated time.
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High Speed Calculations
EXERCISE 2.1
Form this grid and complete without erasing any answer. Check
your time (without any mistake)
Time 75 seconds- 1st try
60 seconds- 2nd try
4
2
9
7
5
6
3
1
8
10
9
4
3
2
5
6
10
8
1
7
SOME PATTERNS
Look at the first digits of table of 9, they are increasing from 0 to 9.
The second digit is decreasing from 9 to 0. So no need to memorize this table just write
the numbers
0 1 2 3 4 5 6 7 8 9
9 8 7 6 5 4 3 2 1 0
There you have the table of nine
By using the famous deductive technology of mathematical learning. Add to all digits in
table of 5 in increasing order starting from top – 2,4,6,8,10,12,14,16,18,20 to get the table
of 7.
19
High Speed Calculations
Here is something for multiplying nine by seven. Count the seventh finger from the right
and bend it:
There are six fingers to the right of the bent finger, and three fingers to the left. So we
have 9 X 7=63.
Some simple facts
•
•
•
To multiply a number by five, multiply a half of that number by ten
When you multiply a number by two, you just add the number to itself
To multiply numbers that differ by two, multiply the number between them by
itself and subtract one.
Interesting property of table of 8
8 x 1=8
8 x 2=16
8 x 3=24
8 x 4=32
8 x 5=40
8 x 6=48
8 x 7=56
8 x 8=64
8 x 9=72
8 x 10=80
8 x 11=88
8 x 12=96
Integer sum
8
1+6=7
2+4=6
3+2=5
4+0=4
4+8=12=1+2=3
5+6=11=1+1=2
6+4=10=1+0=1
7+2=9
8+0=8
8+8=16=1+6=7
9+6=15=1+5=6
8
7
6
5
4
3
2
1
9
8
7
6
Do you see
the pattern,
sequence 9 to
1 repeats.
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High Speed Calculations
Similarly for table of 6
6 x 1=6
6 x 2=12
6 x 3=18
6 x 4=24
6 x 5=30
6 x 6=36
6 x 7=42
6 x 8=48
6 x 9=54
6 x 10=60
6 x 11=66
6 x 12=72
Integer sum
6
1+2=3
1+8=9
2+4=6
3+0=3
3+6=9
4+2=6
4+8=12=1+2=3
5+4=9
6+0=6
6+6=12=1+2=3
7+2=9
6
3
9
6
3
9
6
3
9
6
3
9
Do you see
the pattern,
6,3,9 repeats.
Its an easy
way to
remember.
21
High Speed Calculations
CHAPTER 3
Speeding up Basic Multiplication
Well if the multiplication tables can be memorized, there is nothing more convenient. But
it does remain a bottleneck for most of us and sometimes a miss in the cramming
sequence may lead to disastrous results.
In this chapter, some very nice skills for small number multiplication are being discussed
which help in reducing speed,.
There are rules of multiplication for numbers starting from 3 to 12, but few of them are so
complex that it is better to follow the conventional method.
•
•
The first thing to note is that add a 0 to the left side of the multiplicand.
The second thing is half of an odd number is unit value only, i.e half of 3 is 1,
of 5 is 2 etc.
RULE OF 6
The rule is to each number add half of the right-hand
side digit. In case the digit is odd, add 5 to half of the
right-hand side digit.
Add 0 to the left side of 6892
06892 x 6
2
S T E P S
2 is the first number and it has no neighbour, so write 2
52
9 is second number and odd, hence add 5 and half of 2, i.e. we have
15, write 5, carry over 1
352
8 is third number, add half of 9 which is 4, and the carry over 1, i.e.
8+4+1=13, write 3, carry over 1
1352
6 is the fourth number, add half of 8, i.e. 10 add carry over 1, to get
11, write 1, carry over 1
41352
FINAL ANSWER
IMPORTANT: 0 is to be considered part of number, hence to 0 add
carry over 1 and half of 6, i.e. 3. Write 4
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High Speed Calculations
EXERCISE 3.1
Multiply following numbers with 6 using above rule
Time 30 seconds
1.
2.
3.
4.
RULE OF 7
84622
57793
568903
900864
The rule is to double each number add half of the righthand side digit. In case the number is odd, add 5 to half
of the right-hand side digit.
Add 0 to the left side of 6892
06892 x 7
S T E P S
4
2 is the first number and it has no neighbour, so double 2,write 4
44
9 is second number and odd, double 9 to get 18, add 5 and half of
neighbour 2, i.e. 18+5+1=24, write 4 and carry over 2
244
8 is third number, double 8,add half of 9 which is 4, and the carry
over 2, i.e. 16+4+2=22, write 2, carry over 2
8244
48244
FINAL ANSWER
EXERCISE 3.2
6 is the fourth number, double 6, add half of 8, add carry over 2, i.e.
12+4+2=18 write 8, carry over 1
IMPORTANT: 0 is to be considered part of number, hence to 0 add
carry over 1 and half of 6, i.e. 3. Write 4
Multiply following numbers with 7 using above rule
Time 40 seconds
1.
2.
3.
4.
34662
97093
569763
306884
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High Speed Calculations
RULE OF 5
The rule is to add half of the right-hand side digit. In
case the number is odd, add 5 to half of the right-hand
side digit.
Add 0 to the left side of 6892
06892 x 5
S T E P S
2 is the first number and it has no neighbour, so 0
0
9 is second number and odd, add 5 to half of neighbour 2, i.e.
5+1=6, write 6
60
8 is third number, add half of 9 which is 4
460
6 is the fourth number, add half of 8
4460
34460
FINAL ANSWER
EXERCISE 3.3
IMPORTANT: 0 is to be considered part of number, hence to 0 add
half of 6, i.e. 3.
Multiply following numbers with 5 using above rule
Time 20 seconds
1.
2.
3.
4.
RULE OF 11
34662
97093
569763
306884
This is a slightly different rule.
1. The last number of the number to be multiplied is written
below as right hand digit of answer.
2. Each successive number of the multiplicand is added to its
right side number.
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High Speed Calculations
3. The left hand digit of the multiplicand is placed as the first
digit of answer
06892 x 11
2
12
812
5812
75812
FINAL ANSWER
EXERCISE 3.4
S T E P S
2 is the first number and written as it is.
9 is second number , adding 2 to it gives 11, write 1 and carry over
1
8 is third number, adding 9 and carry over, we have 8+9+1=18,
write 8 and carry over 1
6 is the fourth number, add 8 plus carry over of 1 to get 15, write 5
and carry over 1
IMPORTANT: 0 is to be considered part of number, hence to 0 add
6 and carry over 1 to get 7.
Multiply following numbers with 11 using above rule
Time 20 seconds
1.
2.
3.
4.
RULE OF 12
06892 x 12
4
04
34662
97093
569763
306884
The rule is to add double each digit and add to its right
side neighbour.
S T E P S
double 2 and write 4 as it has no right side neighbour.
double 9 and add 2 , to get 18+2=20, write 0 and carry over 2
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High Speed Calculations
704
double 8, adding 9 and carry over of 2, we have 16+9+2=27, write
7 and carry over 2
2704
double 6, add 8 plus carry over of 2 to get 12+8+2=22, write 2 and
carry over 2
82704
FINAL ANSWER
EXERCISE 3.5
IMPORTANT: 0 is to be considered part of number, hence to 0 add
6 and carry over 2 to get 8.
Multiply following numbers with 12 using above rule
Time 30 seconds
1.
2.
3.
4.
34662
97093
569763
306884
There are some rules for multiplication of 3,4,8 and 9 also, but they are complex and the
conventional method is better.
A nice way to multiply a number by 9 is to write a zero on right side of the number and
subtract the number itself from this.
For example,
42671 x 9, add 0 to 42671, to get 426710, and now
426710
- 42671
384039
EXERCISE 3.6
Using above rules calculate
Time 180 seconds
1.
3.
5.
7.
9.
34567 x 9
149675 x 5
400987 x 12
39753 x 6
900872 x 7
2. 798324 x 11
4. 577439987 x 5
6. 432908 x 11
8. 665778 x 6
10. 75409006 x 12
The next chapter will deal with higher order multiplication techniques. It is advised to
keep practicing till the stipulated time against each exercise is achieved.
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High Speed Calculations
CHAPTER 4
Multiplication of two digits
After you have achieved a certain proficiency in multiplication of small digits, it is now
time to understand certain techniques for multiplying two digits by two digits.
One technique is crosswise multiplication. The Steps are
1. Multiply the left hand digits and right hand digits. Put them down in the extreme
ends respectively.
2. Cross-multiply the digits and put down their sums, carry over to left side digit if
required.
43
x 54
put down 4 x 5 =20
step 1
12 put down 3 x 4=12
step 2
20
31
2322
multiply cross wise
4 x 4 and 3 x 5
total = 16+15=31
step 3
add carry over 1 of 12 to 31, write 2
add carry over 3 to 20 , to get 23
thus the answer is obtained
This may seem a bit complex at the first instance, but after some practice it gets so easy
that you don't even need to write the steps and the correct answer is obtained instantly.
EXERCISE 4.1
Find the products mentally
Time 30 seconds
1.
3.
5.
7.
9.
12 x 13
54 x 67
53 x 67
22 x 11
90 x 56
2. 67 x 77
4. 12 x 14
6. 90 x 80
8. 87 x 12
10. 30 x 89
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High Speed Calculations
NUMBERS CLOSE TO BASE
There are certain cases when the numbers are close to a base. Base means numbers like
10, 100, 1000 etc. There is a very simple way to calculate the product then
The first thing to understand is the COMPLEMENT from a base. We shall be dealing
with only two digit numbers in this chapter; hence for all cases the base will be 100.
Complement of a number from a base is the amount by which the number is less or
more than the base. For 97, complement is -3 as 100-3=97, for 82 complement is -18 as
100-18=82.
RULE
1. Find the complement of the numbers and write it on right side.
2. Multiply the complements and write down in tens and units form, i.e. 2 should be
written as 02.
3. Cross add (since the complement is negative it amounts to subtraction) the
complement from the main number; any number can be used as the answer will be
same.
Following example would illustrate this,
97
-3
100-97=3, complement is 3
x 96
-4
100-96=4, complement is 4
12
Multiply 3 x 4 = 12
97+(-4)or 96+(-3) both are equal to 93,
write down 93
93
9312
Thus 9312 is the answer
Let us take another example,
91
-9
100-91=9, complement is 9
x 86
-14
100-86=4, complement is 14
1
carry
over
26
77
7826
Multiply 9 x 14 = 126, write 26 and 1 is carry over
86-9=77
write down 77, add 1 to 77
Thus 7826 is the answer
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High Speed Calculations
EXERCISE 4.2
Find the products using the complement rule.
Time 30 seconds
1. 98 x 89
3. 89 x 99
5. 99 x 97
2. 88 x 97
4. 95 x 85
6. 94 x 93
Now we shall understand multiplication of numbers that are above the base or one is
below and other above the base.
The Rule is same except that the sign is important in complement and there is a slightly
different way of arriving at the answer. If the complement is for a number less than the
base it is negative written with a dash. In case it is for a number above base, it is written
plainly.
Consider this example
91
-9
91-100=-9, complement is -9
x 106
6
106-100=6, complement is 6
-54
97
00
9646
ATTENTION HERE
Multiply -9 x 6 = -54
91+6=97,
Attention put down two zeroes to
identify as hundreds.
Now add 9700 to -54, or 9700-54
One more example would suffice in understanding this technique
104
4
x 111
11
115
11544
44
4 x 11=44
00
104+11=115
Place 115 and 44 together
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High Speed Calculations
See how simple it becomes, for these particular cases. A definite question in your mind
would be what about numbers that are not close to the base. This of course is one
deficiency of this technique but if the numbers are close to each other there is a way out.
To calculate numbers away from base we can use multiples or sub multiples of the bases.
IMPORTANT TIP
Whatever base is used 10, 100 or 1000, the calculations involving the complements
restrict to number of digits of base. To illustrate, in base of 10, 27 would mean 7 and
carry over 2, but in base 100 it would mean 27 and in base 1000 it would mean 027.
With the above knowledge let us calculate
using base 10
8
x9
7
-2
-1
2
99
x 98
97
-1
-2
02
using base 100
If you have understood what has been done, then proceed further otherwise please go
back to previous sections.
Suppose you were given to find 48 x 47, both base 10 and base 100 will prove more
cumbersome than the direct method. Now this calculation will be illustrated using both
multiple of 10 and sub multiple of 100.
Remember only the left-hand side of the number will have to be multiplied or divided by
the multiple. The right side that is the product of the complements will remain
unchanged.
Let us take the base as 50 = 100/2
48
x 47
45
divide 45 by 2 as 50 =100/2
Answer is 2256
22.5 is actually 22.5 x 100=2250,
add 06 to this
22.5
2256
-2
-3
06
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High Speed Calculations
Let us take another example
41-4=37
41
x 46
37
divide 37 by 2 as 50 =100/2
18.5
Answer is 1886
18.5 is actually 18.5 x 100=1850,
add 36 to this
-9
-4
36
1886
Now using 10 x 5 = 50 base for same numbers
41-4=37
41
x 46
37
write 6 as units digit with 3 as carry over
multiply 37 x 5 as 50 =10 x 5
Add 3 to 185 and get 188, so answer is
-9
-4
36
6
185
1886
Here are some quick examples without full explanation but self evident
base 10 x 7
67
-3
69
-1
66
3
66 x 7=
4323
432
base 10 x 4
47
7
39
-1
46
-7
46 x 4= 1840-7
184
1833
base 10 x 3
31
1
29
-1
30
-1
30 x 3=
900-1
90
899
base 10 x 2
19
-1
18
-2
17
2
17 x 2=
342
34
31
High Speed Calculations
EXERCISE 4.3
Find the products using the above rules.
Time 60 seconds
1. 102 x 104
3. 109 x 87
5. 65 x 72
7. 87 x 91
9. 21 x 25
11. 31 x 28
2. 78 x 79
4. 60 x 58
6. 102 x 89
8. 45 x 48
10. 54 x 53
12. 12 x 16
ALGEBRAIC EXPLANATION OF THE RULE
( x + m) × ( x + n) = x 2 + mx + nx + mn = x( x + m + n) + mn
It is actually so easy to perform multiplication,
x+m+n is what we do in the second step.
m.n is achieved by multiplying the complements.
There is no trick, only a brilliant use of an algebraic expression.
In the next chapter we shall consider cases of higher order multiplication in varying
degrees of complexity.
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High Speed Calculations
CHAPTER 5
Higher Order Multiplication
Now we shall proceed to understand higher order multiplication. The principle used is
same. The base order is 1000 or multiple/sub multiple of 1000 and even higher. Also an
altogether new technique for multiplication of numbers that are not close to each other
will be explained.
Base is 1000
991-14=977
so answer is
991
x 986
977
-9
-14
126
977126
Base is 7 x 100
706
x 691
691+6=697
This step is important 697x 7= 4879
Answer is
6
-9
-54
487900
(as base is 100)
487846
Some quick examples, please observe carefully. The details are purposefully hidden so
that you can understand yourself.
base 100 x 5
467
-33
489
-11
456
3/63
456 x 5
=2280 228363
base 1000 / 2
509
9
497
-3
506
-27
506/2 253000
=253
-27
=252973
base 100 x 7
147
-553
689
-11
136
60/83
136 x 7=
95200
952
+6083
=101283
Still simpler !
base 100 x 5
499
-1
18
-482
17
482
17 x 5=
8982
85
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High Speed Calculations
base 100 x 6
489
-111
612
12
501
-1332
501 x 6 = 300600
3006
-1332
=299268
base 100 x 5
499
-1
18
-482
17
482
17 x 5=
8982
85
base 100 x 9
890
-10
918
18
908
-180
908 x 9= 817200
8172
-180
=817020
base 100 x 3
303
3
312
12
315
36
315 x 3 =
94536
945
It is seen that the products above are all close to each other, except for a few cases where
the second number is small and just a little bit of lateral thinking enabled us to simplify
the calculation.
EXERCISE 5.1
Find the products
Time 90 seconds
1.
3.
5.
7.
9.
912 x 913
544 x 567
293 x 307
220 x 211
390 x 406
2. 667 x 707
4. 812 x 813
6. 907 x 880
8. 898 x 12
10. 309 x 289
Although the conventional technique of multiplication is acceptable and reliable but the
following technique when mastered can turn the reader into a master in calculation
without using a line more than just the answer. But this requires considerable practice
and perfection as one error could lead to disastrous results and a lot of embarrassment.
Let us begin with the smallest example. This method is only a reorder of the
conventional technique, but is extremely useful in reducing the time.
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High Speed Calculations
Vertically and Crosswise multiplication
The above diagram depicts the procedure for vertically and crosswise multiplication.
•
•
•
•
•
•
The two numbers should be written one above the other. Start from the right side
of the number, which is the 5th step in the diagram.
Multiply both the right side members and write down the unit digit and carry
over.
Now multiply crosswise as the 4th step and add the products, write down the unit
digit and carry over the ten's digit.
Now look at the 3rd step in the figure, crosswise multiply the extreme end digits
and then add the vertical product of the middle digits.
1st and 2nd steps are self-evident.
This is the one line product; remember to add the respective carryovers to the left
side neighbouring digit.
The following example would clearly illustrate this
3 5 7
x 7 3 2
1st step 2 x 7 = 14
2nd step 2 x 5 + 3 x 7 = 31
21 4 0 1 4
4...7...3...1.....
carry overs
3rd step 3 x 2 + 7 x 7 + 5 x 3 = 70
4th step 3 x 3 + 5 x 7 = 44
5th step 3 x 7 = 21
2 6 1 3 2 4 Final result after
adding carryovers
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High Speed Calculations
Let us take another example
2 9 6
x 4 5 7
1st step 6 x 7 = 42
8 6 3 3 2
2nd step 9 x 7 + 5 x 6 = 93
4...8...9...4.....
carry overs
3rd step 2 x 7 + 6 x 4 + 9 x 5 = 83
4th step 2 x 5 + 9 x 4 = 46
5th step 2 x 4 = 8
EXERCISE 5.2
1 3 5 2 7 2 Final result after
adding carryovers
Find the products in one line. Do all the additions mentally
Time 180 seconds
1.
3.
5.
7.
9.
678 x 346
145 x 768
674 x 350
452 x 469
190 x 781
2. 294 x 498
4. 380 x 189
6. 987 x 876
8. 961 x 798
10. 129 x 459
36
High Speed Calculations
CHAPTER 6
Simple Division
To understand the technique of a division, we shall first look at the following examples of
division by nine. Before we proceed let us remind us of the terms,
Dividend
Divisor
Quotient
Remainder
this is the number which is to be divided
this is a number which would divide.
the whole number whose product with divisor is closest to dividend.
the remaining part of the dividend after division.
divisor
)
(
dividend
quotient
______________
remainder
dividend = (divisor x quotient) + remainder
AN INTERESTING PROPERTY
From the Ex 1 below⇒ 12 = (9 x 1) + 3
9 1/ 2
9 2/1
9 7/0
9 6/2
.. / 1
.. / 2
.. / 7
.. / 6
1/ 3
2/3
7/7
6/8
Ex 1
Ex 2
Ex 3
Ex 4
The first step is to divide the dividend in part with a / (slash). An interesting property
emerges in the dividend.
The left part becomes the quotient and sum of the two numbers becomes the
remainder.
Consider Ex 1; 12 divided by 9, place a slash between 1 and 2. Now write 1 below 2,
add them to get 3. Write it below the line and then copy 1 below the line with a slash
in between. Thus 1 is the QUOTIENT and 3 is the REMAINDER
Ex 4; 62 ÷ 9, write 6 below 2 , add 2 + 6 = 8 , thus 6 is quotient and 8 remainder
Using this basic property the division rules shall be deduced. Let us also see if the same
property works for higher order numbers. The point to be kept in mind here is the left part
37
High Speed Calculations
of the slashed dividend has to be divided by 9 and then written below the original
dividend.
In this case split the dividend into two parts, since a two digit number 20
is the left part, using the previous method, write down quotient and
remainder of 20 below the number with a usual slash as depicted. Add up
the right side and left side to get the final result.
9 20 / 5
..2 / 2
22 / 7
One more example
In this case split the dividend into two parts, since a two digit number
14 is the left part, using the previous method, write down quotient and
remainder of 14 below the number with a usual slash as depicted. Add
up the right side and left side to get the final result.
9 14 / 7
..1 / 5
15 / 12
16 / 3
But here as you must be seeing 12 can certainly not be a remainder,
divide 12 by 9 and add 1 to 15 and 3 is the remainder. So final answer is 16 quotient
and 3 remainder.
The general rule is if the remainder is same or greater than the divisor,
redivide the remainder by 9, carry the quotient over to the quotient
column/side and retain the result which is final remainder.
RULES FOR DIVISION BY NEAR TO BASE METHOD
1. Decide on the base, for two digit numbers it is 100, three digit numbers
1000.
2. Find the complement of the divisor from this base i.e. for 89, 11 is the
complement from 100. Similarly for 996, 4 is the complement from 1000.
3. Place a slash between the dividend depending on the base, thumb rule is
slash should be placed as per the number of zeroes in the base, i.e. if base
is 1000, place slash three digits from right.
4. Multiply the complement with quotient and follow the same rules.
8 4/7
2 is the complement of 8
.. / 8
Thus 4x2=8 is written below 7
4 / 15
5/7
But 15 is higher than 8, the divisor, so redividing it by 8 we get 5 as
quotient and 7 as remainder.
A better way to remove confusion during division is to write the complement below the
divisor. Let us look at another example.
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High Speed Calculations
73
1 / 15
27
.. / 27
1 / 42
To divide 115 by 73
Write 27 the complement of 73 from 100
Place a slash, two digit from right
Multiply 27 x 1 and add to 15.
Let us look at some more examples.
888
1 / 149
112
.. / 112
7998
1 / 3618
2002
.. / 2002
993
2 / 159
7
... / 014
1 / 261
1 / 5620
2 / 173
EXERCISE 6.1
Using the rules explained above, solve the following
1.
3.
5.
7.
9.
678÷9
119÷89
14567÷8998
32345÷8888
394567÷89997
2. 254÷9
4. 2245÷997
6. 5467÷887
8. 4532÷991
10. 447÷86
All the above examples would have made the basic method clear. Now we shall seek to
understand division when the dividend is much larger than divisor and there is more than
one digit on the left side of the slash.
Divide 106 by 7
Complement of 7 , 3 is written below 7
7 ) 10 /6
3
7 ) 10 /6
3 ____
1 /
7 ) 10 /6
The first digit of the quotient,1 is multiplied by
3
3
complement, 3 and answer is placed below the next
____
dividend digit
1 /
7 ) 10 /6
The second column is added up 0+3=3, this is the
3
3
second quotient digit.
____
13 /
The first digit of the dividend is brought down to
the answer
39
High Speed Calculations
The Final step is to multiply this new quotient digit
by the complement, i.e. 3x3=9, which is placed
after the slash. This column is added up to get the
remainder.
But 15 obviously cannot be the remainder, when
divided by 7 it gives 2 and remainder 1, add this
quotient. Thus the final result is 15 as quotient and
1 as remainder.
7 ) 10 /6
3 3 /9
____
13 /15
7 ) 10 /6
3 3 /9
____
15 /1
Let us take another example
Divide 10113 by 88
Complement of 88 , 12 is written below
The first digit of the dividend is brought down to
the answer
The first digit of the quotient,1 is multiplied by
complement, 12 and answer is placed below the
next dividend digit
The second column is added up 0+1=1, this is the
second quotient digit.
This second quotient digit is again multiplied by
complement, 12 and written below one digit being
after the slash.
The third quotient digit is now multiplied by
complement and written below the next two digits
of the dividend. Add up all the right side numbers
for remainder
REMEMBER the partial result is placed by
shifting towards the right side by one digit after
each step.
88 ) 101 / 13
12
88 ) 101 / 13
______
1 /
88 ) 101 / 13
12
______
1 /
88 ) 101 / 13
12
______
11 /
88 ) 101 / 13
12
1/2
______
114 /
88 ) 101 / 13
12
1/2
/ 48
______
114 / 81
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High Speed Calculations
Let us look at another simple example, try to figure out what has done
8997
20 / 1151
1003
..2 / 006
.... / 2006
22 / 3217
•
•
•
•
•
•
1003 is the complement
Slash is placed 4 digits from right as
10000 is the base
1003 x 2 = 2006 is written first
0+2 is the second digit of quotient
Again 1003 x 2 = 2006 is written
shifting the one digit towards right
Adding up for remainder
There is one very important thing to understand and the following example would clearly
illustrate this. The reader may get confused on what to do when the addition results in
more than one digit in quotient and when it is more than one digit in remainder.
4363÷9
4 is the 1st digit of quotient
4x1=4 , 4+3=7 is the 2nd digit
7x1=7 , 6+7=13 is the 3rd digit.
IMPORTANT Here the conventional rule applies for
quotient. Carry over 1 to next column and 2nd digit
becomes 7+1=8
But careful for the remainder 13x1=13 is placed after the
slash and remainder become 3+13=16 and not 3+13=43
Here the rule is different, 16 should be redivided by the
divisor 9 to get 7 as remainder and 1 is added to quotient
digits 483 to get 484 as result
EXERCISE 6.2
9)4 3 6/ 3
4 /
7/
/13
¯¯¯¯¯¯¯¯¯¯¯
4 7 13 /16
4 8 3/ 7
1
¯¯¯¯¯¯¯¯¯¯¯¯
4 8 4/ 7
Final result
Using the rules explained above, solve the following
1. 23456÷89
3. 125467÷887
5. 121679÷898
2. 121237÷997
4. 314578÷987
6. 2104569÷9892
One very obvious question would be what about numbers above the base like 11, 109,
1007 etc. For such numbers method is same except that the complement is actually
surplus and is called surplus. For 11, surplus is 1 as it is 1 above 10 the base.
High Speed Calculations
41
After all this while, you must be wondering that these rules are brilliant but only for very
specific cases where the divisor is close to some base. There is another rule, which can
substantially reduce time for division by any number irrespective of the base. In
competitive examinations and normal school examinations, the level of difficulty in
division is not very high and commonly it is these small two digit or three digit divisors
that pose a problem. The following technique in combination with above techniques
provides a very simple and safe system for division. You only have to add or subtract and
not be involved with big multiplication, which is another headache.
This new technique will be explained in the next chapter.
42
High Speed Calculations
CHAPTER 7
More Division
The general method for division taught conventionally is called the straight division
method. It covers all possibilities. There is one superb way of easing calculations with
some mental arithmetic and no big multiplication etc.
Let us start with a very simple example of 467÷32
2
First step is to set out is the first digit of the divisor and
write it on top of the remaining number, in this case 2
called the flag digit is written on top of 3. The remainder
stroke will be placed one digit from right, as the number
of FLAG digits is also one.
3
4 6/7
2
Now only 3 shall be used for dividing.
3
4÷3 gives us quotient 1 and remainder 1.
Place the quotient below the line as the first digit of main
quotient. Place the remainder between 4 and 6 in the next
line.
IMPORTANT
Using the flag digit 2, subtract from 16 the product of this
2
flag digit 2 and the previous quotient 1.
3
16 - {2(flag digit) x 1 (previous quotient)} =14
This is now the new dividend. Divide 14 by 3 to get
quotient 4 and remainder 2.
Place them as before 4 below the line and 2 between 6
and 7.
2
We are now left with the remainder portion. Here we do
3
not divide but merely find the remainder by subtracting
the product of previous quotient digit and flag digit.
27 - {2(flag digit) x 4 (previous quotient)} =19.
This is the final remainder.
4 6/7
1
1
4 6/7
1 2
1 4
4 6/7
1 2
1 4 / 19
It is heartening to find that without much to multiply and simple addition and subtraction
with a set of rules we can divide any number by any number. There is a simple algebraic
proof behind this, which is not very important to know as long as the procedure is
understood.
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High Speed Calculations
SUMMARISING THE RULES FOR STRAIGHT DIVISION
9 Set out the flag digit. The flag digit is obtained by removing the first digit of the
divisor. For example, for 35 flag digit is 5 and in case of 367 flag digit is 67, after
removing 3.
9 The number of the flag digits indicates the number of digits to be placed after the
remainder stroke.
9 The first digit of the divisor, called modified divisor, does the dividing.
9 The first digit of the quotient is obtained by dividing the first number of the
dividend by the modified divisor.
9 This quotient be placed below the line and the remainder just before the next
number of the dividend.
9 The new dividend is obtained by combining the remainder with the next digit,
subtract from this digit the product of the flag digit and the previous quotient.
9 Repeat this procedure till the / mark.
9 The remainder is obtained by subtracting the product of the flag digit and the
immediate quotient from the combined number.
Another example would depict the procedure clearly; we shall use a single line.
Find 98613 ÷ 63
9 First find the flag digit which is 3
9 Mark the slash, one digit from right side between 1 and 3
3
9
6
1
Steps
8
3
9÷6
q=1
r=3
6
5
1
3
3
5
6
5
38-(3x1)
=35
35÷6
q=5
r=5
56-(3x5)
=41
41÷6
q=6
r=5
51-(3x6)
=33
33÷6
q=5
r=3
here, q depicts quotient and r is remainder.
/
5
/
18
Remainder
column
33-(3x5)
=18
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High Speed Calculations
Picking up another example. 178291÷83
9 The flag digit is 3
9 Slash to be placed before 1
3
8
1
7
8
1
2
Steps
EXERCISE 7.1
1.
3.
5.
7.
9.
17÷8
q=2
r=1
2
9
4
/
7
1
3
1
4
8
18-(3x2)
= 12
12÷8
q=1
r=4
42-(3x1)
=39
39÷8
q=4
r=7
79-(3x4)
=67
67÷8
q=8
r=3
/
7
Remainder
column
31-(3x8)
=7
Using the rules explained above, solve the following
423÷21
4911÷32
8466÷64
86369÷76
34632÷52
2. 651÷31
4. 1668÷54
6. 59266÷53
8. 38982÷73
10. 59266÷83
ALTERED REMAINDER
Observe the following division using this method
6013÷76
6
7
6
0
1
4
8
Steps
60÷7
q=8
r=4
41-(6x8)
=41-48
which is
-7
/
3
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High Speed Calculations
It is found that that in the next step the result is negative. This would be happening many
times during divisions.
For cases like these the altered remainder is used, which is simply like this
60÷7 q=8, r=4
or
q=7, r=11 here 7x7=49 and 60-49=11
This is called alteration of remainder.
The quotient is reduced by 1, and resulting remainder is picked for further division.
Now we shall proceed with the same problem using the altered remainder.
6
6
7
0
1
/
11
7
Steps
9
/
111-(6x7)
=111-42
= 69
69÷7
q=9
r=6
60÷7
q=7
r=11
3
6
9
Remainder
column
63-(6x9)
=9
Note : quotient is reduced from 8 to 7 in
first step and new remainder 11 is carried
over to form next number 111
Another example,
6
6
1
5
2
Steps
7
3
15÷6
q=2
r=3
0
/
7
3
37-(6x3) =25
25÷6,
normally
q=4, r=1
but r has to
be modified
q=3,r=7
7
10
7
70-(6x3) =52
52÷6,
normally
q=8, r=4
here also r
needs to be
modified
q=7,r=10
/
65
Remainder
column
107-(6x7)
=65
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High Speed Calculations
EXERCISE 7.2
1.
3.
5.
7.
9.
Using the rules explained above, solve the following
3412÷24
1408÷23
4919÷11
9678÷59
9836÷21
2. 8218÷33
4. 4143÷18
6. 5711÷54
8. 6712÷53
10. 601325÷76
DIVISION BY TRANSPOSE AND ADJUST
Using the Remainder Theorem And Horner's Process Of Synthetic Division, one very
easy method for division of numbers close to bases has been devised.
Before we proceed let us review some concepts
Surplus and deficient.
Base 10
11 has
8 has
1 surplus
2 deficient
Base 100
109 has
89 has
9 surplus
11 deficient
For this method surplus is prefixed with – as it has to be subtracted.
The principle behind this is GET TO BASE.
To get 11 to base i.e. 10, we have to add -1 to 11; 11+(-1)=10
Similarly to get 98 to base i.e. 100, we have to add 2 to 98; 98+2=100
Let us start with a very simple example,
1467÷111
here the divisor 111 has 11 surplus, so this is written as -1 -1 below 111.
Since the base is 100 hence slash is marked before the two digits from right.
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High Speed Calculations
PROCEDURE
The method is same as explained in chapter 6, here also the complement/surplus is
written below the divisor.
•
•
•
•
•
The left most digit is written below the line
The surplus is multiplied by this partial answer and placed below from the second
digit from left onwards (please see the example below)
The second part is added up to get the partial quotient and the surplus is
multiplied with this result.
The process is carried on till the last digit is covered.
The normal rules for addition are used and the quotient and remainder are the
numbers left and right of slash respectively.
The following examples will explain the method and some special cases.
1 1 1
-1 -1
1
4
-1
1
/ 6
-1
-3
/ 2
3
7
-3
4
Thus 13 is the quotient and 24 the remainder.
Another example, 16999÷112
1 1 2
-1 -2
1
1
1
6
-1
5
5
9
-2
-5
/ 9
9
2
-10
-2
/ -3
-4
5
-1
+11
2
1
/ 8
7
Here a strange case has emerged, negative remainder. There is a simple rule; it actually
means we have over divided the dividend. Therefore reduce the quotient by 1 and add the
divisor to the remainder.
Reduce 152 by 1 to have 151
Add 112 in the remainder column as shown, +11 with first digit and +2 with second digit.
Thus answer is q = 151 and r = 87.
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High Speed Calculations
Let us take another example,
1 1 3 2
-1 -3 -2
1
st
2
-1
1
1
nd
2
rd
1
3
1
/ 3
-3
-1
/ -1
4
-2
-3
-1
9
-2
7
-1
+11
+3
+2
0
/ 10
2
9
Here again remainder had two negative digits, reduce 11 by 1 to get 10.
Add +11, +3and +2 to remainder columns respectively.
Thus q =10 , r =1029
Isn't this wonderful. Although this technique applies to specific cases but in most of the
competitive examinations, the question setter ensures that the division remains as simple
as possible. And for most of the cases these techniques would really help in reducing time
for calculation.
EXERCISE 7.3
1.
3.
5.
7.
9.
Using the rules explained above, solve the following
1234÷112
1234÷160
12789÷1104
21188÷1003
15542÷110
2. 1241÷112
4. 239479÷11203
6. 27868÷1211
8. 25987÷123
10. 189765÷1221
DIVISION BY TRANSPOSE IF NUMBERS ARE LESS THAN BASE
Let us consider the number 819, the deficiency is 181, which means we use + sign and
division by transpose and adjust method can be done.
8 1 9
1 8 1
1
1
/ 4
1
/ 5
6
8
14
7
1
8
1
/ 6
4
8
This case was simple but big number 8 is involved. this can be simplified by judicious
use of surplus and deficient. 819 is 1000 - 181 or (200-20+1), thus 181 can be written as
2 -2 1
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High Speed Calculations
8 1 9
2 -2 1
1
/ 4
2
/ 6
1
6
-2
4
7
1
8
Let us see another example
49999÷9891
in this case 9891 is 10000-109 and 109 can be expressed as (100+10-1) or 1 1 -1, easing
the calculation tremendously. That is a bit of lateral thinking.
9 8 9 1
0 1 0 9
0 1 1 -1
4
/ 7
0
9
4
9
4
9
-4
st
4
/ 7
13
13
5
nd
4
/ 8
4
3
5
1
2
IMPORTANT: The number of operators is same as number of zeroes in the base.
For example in the previous example if you start working with only 1 1 –1 ignoring the
0 the whole result would be wrong. Be careful in such cases.
Notice that in 3rd and 4th column the additions result in larger than 10 numbers. The rule
is add the carry over to the preceding column number starting from right side. But
remember each column is a unit, ten, hundred and so on, so the addition has to follow the
same rule.
For example if 13 of 3rd column is to added by carry over 1 from 4th column, it will be
like this
13
1
IMPORTANT: Consider each column separately do not add in the conventional
manner.
Suppose some addition leads to a negative number, then the result should be subtracted
from preceding column considering its place value i.e adding a zero to it. For example if
in 3rd and 4th column results in remainder row are 3 and -4 respectively, then the final
column would result in 30-4 = 26 and not -1, because 3rd column's place value is 10 more.
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High Speed Calculations
To illustrate this let us consider another example,
Here 8829 = 10000 - 1171 and 1171 can be expressed as (1000+200-30+1) or 1 2 -3 1
8 8 2 9
1 1 7 1
1 2 -3 1
2
/ 5
2
4
4
5
-6
3
2
2
/ 7
8
-1
5
7000
800
-10
+5
5
Here -1 has to added to 8 x 10 = 80,
This remainder row is actually
2
/ 7
7
9
Answer is q = 2 and r = 7795
EXERCISE 7.4
1.
3.
5.
7.
9.
Using the rules explained above, solve the following
13044÷988
31883÷879
10091÷883
39979÷9820
134567÷8893
2. 7101÷878
4. 21002÷799
6. 2683÷672
8. 20121÷818
10. 24434÷987
The reader must be wondering on how to choose a particular method. It can be very
confusing to first learn all the techniques and then remember when to use what. One very
useful way to know which technique to use when is,
Check whether the divisor is close to a base. The technique explained in Chapter 6 is
actually the same as Division by transpose and adjust. The only difference is the method
adopted. Sometimes the method in Chapter 6 is easier as it is direct. But for higher order
numbers the technique explained in this chapter is better.
If the number is not close to the base use the direct division method, which can be applied
to any set of numbers.
Let us do an exercise of choosing the method to be adopted.
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High Speed Calculations
EXERCISE 7.5
1.
3.
5.
7.
9.
By visual inspection choose the method for division.
13044÷988
39883÷29
401÷88
39979÷1003
134567÷19
2. 7101÷78
4. 21002÷457
6. 7683÷78
8. 60121÷1008
10. 54434÷563
Happy dividing. The best way to decrease the time for calculating is not to try to increase
speed. Practice more such problems by making your own problems and checking the
answers by reverse method of multiplying the divisor with quotient and adding the
remainder to get the dividend.
DIVISOR x QUOTIENT + REMAINDER = DIVIDEND
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High Speed Calculations
CHAPTER 8
Quick Squaring
The multiplication techniques explained in Chapter 4 are employed for quick squaring
with a slight modification. In most of the competitive examinations the squaring
problems are usually not very complex, but calculating the same does take time, which
can be reduced by 50% using the techniques, which will be explained now.
All the procedures lead to an increase in calculation speed once adequate practice has
been done. The reader may now wonder on how to remember the techniques. Case to
case examination will very smoothly orient your mind and the choice would be a reflex
action, rather than a very conscious effort.
FOR NUMBERS NEAR TO A BASE
For numbers near to base
1. First find the complement of number from Base, for 97, complement is 3, 100-3,
this can be marked as -3, for 1009, complement is 9 from base 1000.
2. Now "add" this complement to the number, i.e. add -3 to 97, to get 94 or add 9 to
1009 to get 1018.
3. Square the complement and place to the right side of the added result. Here it is
important to note the base value will decide the number of digits on right side, e.g.
in case of 97, complement 3 is squared to get 9 but 9 shall be written as 09 since
we are using base 100 which has 2 zeroes.
The following examples will illustrate this technique,
Squaring of 97
97
-3
97+(-3)
9
94
09
9409
100-97=3, complement is 3
Add -3 to 97
Square -3, add 0 before 9 to make it 2 digit
since we are dealing with base 100
Thus 9409 is the answer, very straight
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High Speed Calculations
Another example,
Squaring of 1009
1009
9
1009+9
81
1018
081
1000-1009= -9 complement is 9
Add 9 to 1009
square 9
add 0 in front of 81 to make it 3 digit
1018081 is the answer
Stepwise explanation was given so that you could understand the procedure, a few
examples are solved in one line for illustration (try them once yourself).
982 = 98-2/04=9604
862 = 86-14/196=72+1/96=7396 (because we are operating in base 100)
9892 = 989-11/121=978121
1042 = 104+4/16=10816
10152 = 1015+15/225=1030225
792 = 79-21/441=58+4/41=6241
9652 = 965-35/1225=930+1/225=931225(because we are operating in base 1000)
1132 = 113+13/169=126+1/69=12769(because we are operating in base 100)
99912 = 9991-9/0081=99820081
999892 = 99989-11/00121=9997800121
EXERCISE 8.1
Find the squares of following numbers
1. 97
2. 108
3. 115
4. 998
5. 899
6. 989
7. 9987
8. 9979
9. 10032
10. 1023
11. 1012
12. 99997
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High Speed Calculations
FOR NUMBERS NOT NEAR TO A BASE
For numbers which are not near to base, the base can be chosen as multiple or
sub multiple of a decimal base.
For 47, 50=100/2 or 50=10 x 5 can be the base.
For 509, 500=100 x 5 or 100/2 can be the base.
Let us now see how this works.
Suppose you were to find 492. Taking base as 50 which is 100/2
Squaring of 49
Take 50 = 100/2 base
49
-1
50+(-1)=49
49+(-1)
01
Add -1 to 49
Square 1 , write as 01
48
48/2
01
IMPORTANT: divide the first part
by 2 to get final answer
2401
It is also possible to solve this by using 50=10 x 5
Squaring of 49
49
-1
49+(-1)
1
48
48x5
1
2401
Take 50 = 10x5 base
50+(-1)=49
Add -1 to 49
square 1 write as 1
IMPORTANT: multiply the first
part by 5 to get final answer
Same as above
By now it must be clear to you that the first part of the sub result has to divided or
multiplied depending on base decided as multiple or sub multiple of nearest decimal base.
A few more examples,
322 = 32+2 / 4=34x3/4=1024 (choosing 10x3 as base)
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High Speed Calculations
192 = 19-1 / 1=18x2/1=361 (choosing 10x2 as base)
582 = 58-2 / 4=56x6/4=3364 (choosing 10x6 as base)
3872 = 387-13 / 169=374x4+1/69=149769 (choosing 100x4 as base)
4982 = 498-2 / 004=496÷2/004=248004 (choosing 1000÷2 as base)
2532 = 253+3 / 009=256÷4/009=64/009=64009 (choosing 1000÷4 as base)
20212 = 2021+21 / 441=2042x2/441=4084441(choosing 1000x2 as base)
49962 = 4996-4 / 016=4992x5/016=24960016 (choosing 1000x5 as base)
49972 = 4997-3 / 0009=4994÷2/0009=24970009 (choosing 10000÷2 as base)
69922 = 6992-8 / 064=6984x7/064=48888064 (choosing 1000x7 as base)
EXERCISE 8.2
Find the squares of following numbers
1. 68
2. 208
3. 315
4. 298
5. 699
6. 789
7. 7987
8. 6987
9. 20032
10. 8023
11. 6012
12. 89997
SQUARING USING DUPLEX
There is yet another method for squaring using the concept of Duplex. In this method the
basic formula,
(a+b)2=a2+2ab+b2
must be remembered.
The Duplex is calculated using two basic rules,
1. For a single central digit, square the digit, a2
2. For even number of digits or a and b equidistant from ends double the cross
product, 2ab
Don't worry if you haven't understood, the following examples will illustrate this,
2 is a single digit, D=22=4
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High Speed Calculations
43 is double digit, D=2 x (4 x 3)=24
313 is three digit, D= 12 + 2 x (3 x 3), as 1 is the central digit and 3 and 3 are
equidistant from the ends. D= 19
5314 is four digit, there is no central digit, thus two numbers in center and at the
ends are used thus, D= 2 x (5 x 4) + 2 x (3 x 1)=46
13451 is five digit, 4 is central digit thus D= 42 + 2 x (1 x 1) + 2 x (3 x 5)=48
Coming back to the method of squaring.
Note: The Square of a number containing n digits, will have 2n or 2n-1
digits.
So, first mark extra dots or space one less than the number of digits for the number
to be squared, e.g. for 456 mark 2 dots, for 45678 mark 4 dots.
The Duplex, D has to be calculated for each part of the number to be squared starting
from the right side. That is if the number to be squared is 12345, then D has to be
calculated using the method explained above for 5, then 45, then 345, then 2345, then
12345 and then 012345 so on till the number of dots which are placed with assumed
value of 0.
Getting confused, it is actually simple start from right side, pick up the digits one by one
get their D and write them down with the units place on main line and the other digits a
line below as depicted.
2232 , add 2 dots
D for 3 = 9
..223
D for 23 = 2x2x3 = 12
D for 223 = 4 + 2x2x3 = 16
D for 0223 ( considering dot as 0) = 2x0x3 + 2x2x2=8
D for 00223 = 22 + 2x0x3 + 2x0x2 = 4
All this may seem very cumbersome but it is not. All these calculations can be done
mentally and there is no need to write all this. This is being done only to explain the
method.
The method is to write the D's from right side like this, keeping the units digit on the
main line and carryovers as subscript.
4 8 16 12 9,
now add the "carryovers" to the immediate left side digit,
Thus 2232 = 49729
Let us now solve a few examples in one line,
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High Speed Calculations
3352 = ..335 ⇒ 9 18 39 30 25⇒ adding up we get 112225, so easy
6782 = ..678 ⇒ 36 84
145 112 64⇒
459684
22342= …2234 ⇒ 4 8 16 28 25 24 16⇒ 4990756
To understand the method better, solve the above yourself, and check at each stage. Do
not haste. No time is saved if you consciously try to save time in calculations.
Concentrate step wise on the problem. Remember this is the shortest possible method
and no more shortcuts are possible.
A few more squared numbers,
3472 = ..347 ⇒ 9 24 58 56 49 ⇒ 120409
4132 = ..413 ⇒ 16 8 25 6 9 ⇒ 170569
7732 = ..773 ⇒ 49 98 91 42 9 ⇒ 597529
21362 = …2136 ⇒ 4 4 13 30 21 36 36 ⇒ 4562496
3952 = ..395 ⇒ 9 54
2
111 90 25
849 = ..849 ⇒ 64 64
⇒ 156025
160 72 81
⇒ 720801
2
3247 = …3247 ⇒ 9 12 28 58 44 56 49 ⇒ 10543009
534912 = ….53491 ⇒ 25 30 49
114 80 78 89 18
1 ⇒ 2861287081
2
4632 = …4632 ⇒ 16 48 60 52 33 12 4 ⇒ 21455424
76932 = …7693 ⇒ 49 84
162 150 117 54
9 ⇒ 59182249
In all the above cases
D the Duplex has been calculated at each stage
Digits other than the unit's place have been subscripted
And then added to left side main digit.
Please do all the above squaring yourself once before proceeding further. After this you
would become an expert at squaring and it would really be appreciated how easy it has
become, and you are a ONE LINE Expert.
EXERCISE 8.3
1. 68
4. 298
7. 7141
10. 8617
Find the squares of following numbers using D method
2. 208
5. 653
8. 6507
11. 5538
3. 315
6. 789
9. 21313
12. 81347
High Speed Calculations
SOME SPECIAL SQUARES
For Numbers ending with 5
There is a special case when the number ends in 5.
Put square of 5 as last two digits. Add 1 to the tens digit and multiply with original.
Take 45, place 25 and then 4 x (4+1) = 4 x 5 = 20 , thus we have 2025
125⇒ 12 x 13 ⇒ 156 ⇒ 1252 = 15625
75⇒ 7x8 ⇒56 ⇒752 = 5625
For Numbers starting with 5
Square the last digit and place the squared quantity as last two digits.
The first two digits are 25 + the last digit
Taking 56 as example⇒ 62=36, place 36 as last two digits
25+6= 31, 3 & 1 are first two digits, thus 562=3136
532= 25+3⇒ 2809 (09 is square of 3)
592= 25+9⇒ 3481 (81 is square of 9)
58
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High Speed Calculations
CHAPTER 9
Square Roots
There are certain rules about square roots of numbers, which are listed below
ƒ
The number of digits in square root of the number will be same as the number of
two digit groups in the number. This actually means if the number contains n
digits, then the square root will contain n/2 or (n+1)/2 digits.
ƒ
The exact square cannot end in 2,3,7 or 8.
ƒ
1,5,6 and 0 as the last digits of the number repeat themselves in the square of the
number.
ƒ
1 and 9, 2 and 8, 3 and 7, 4 and 6, 5 and 5 have the same ending.
A number cannot be an exact square if
ƒ
if it ends in 2,3,7 or 8
ƒ
if there are odd number of zeroes at the end
ƒ
if its last digit is 6 but second last digit is even.
ƒ
if the last digit is not 6 but second last digit is odd.
ƒ
the last two digits of even number are not divisible by 4.
To start with, first make 2 digit groups of the number this will give the number of digits
in square root.
ƒ
Now take the right side digit group to find the first digit of the square root. The
number whose square is closest and lower to this digit group is the first digit.
For example if the digit group is 35 then 25 is lower closest to 35, hence 5 is the
first digit.
ƒ
Find the difference between the first digit group and square of first digit of square
root. This difference should be prefixed to the next digit as was done during the
division process in a line below. This now becomes the new quotient.
ƒ
Double this first digit and place as dividend.
ƒ
Start the division process and find q and r. Place q as the new digit and prefix r to
the next digit of the number.
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High Speed Calculations
ƒ
Be careful in the next step, the new quotient should be reduced by D,the duplex
of the right side of the partial square root. Now divide and place q as new digit in
square root and r as the next digit.
ƒ
In the next step take the duplex of the last two digits, this process is continued till
the last digit of the expected square root is reached. After this the D is calculated
in decreasing order and after reaching the last digit of the number D of the last
digit of the expected square root is used.
ƒ
In fact this difference should be 0 for a perfect square after last digit of the
square root. It is proof of the correct answer.
ƒ
You may also need to alter the remainders as was done in straight division.
All this may be very confusing, but observe the following examples carefully.
Remember the Duplex D, the same shall be used here (refer chapter 8)
2
:
3
1
2
1
5
st
Steps
1 is the 1
digit,
thus 2
becomes the
dividend
2
2-1 =1
4
:
3
using altered
remainder
13÷2, q=5,
r=3
0
3
3
D of 5 is 25,
so
34-25=9,
9 is the new
dividend,
9÷2, here
again using
altered
remainder
q=3, r=3
:
9
0
0
D of 53 is 30,
30-30=0
0
D of 3 is 9,
9-9=0
This can be very confusing, but remember the square root of a number will have n/2
or (n+1)/2 digits, i.e. for a 5 digit number it will have 3 digits, and also for a 6 digit
number 3 digits.
Let us take another exercise of finding square root of 20457529, here n= 8, thus number
of digits in square root will be 4.
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2
0
:
4
5
4
8
4
4
5
st
4 is 1
digit,
so 8 is
dividend
2
20-4
=4
:
7
4
2
:
3
3
:
2
1
0
9
0
0
12-D(23)
=12-12
=0
35D(523)
=35-34
=1
1÷8
q=0
r=1
47-D(52)
=47-20
=27
27÷8
q=3
r=3
45-D(5)
=45-25
=20
20÷8
q=2
r=4
44÷8
q=5
r=4
5
q=0
r=0
0
9-D(3)
=9-9=0
q=0
r=0
By now the use of D must be clear, start with single, then double, triple so on and again
recede form higher to lower digits such that the last division is done with single digit D.
Some more square roots (without details of calculation) are elucidated, do it yourself and
then compare the calculations. It will serve to remove all doubts.
3
0
:
1
5
10
5
7
:
4
11
2
0
8
1
8
4
9
:
0
0
1
8
:
2
4
3
4
:
7
6
10
8
:
6
0
0
Note: Altered remainders are used here
3
3
:
6
4
8
10
5
6
8
:
0
It may be observed that all partial results will be zero after the square root is obtained;
this is verification of the result.
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7
4
5
:
0
2
7
:
6
5
8
8
1
:
7
6
6
8
4
2
0
6
4
:
6
:
5
:
:
3
3
6
:
:
0
0
2
9
4
0
0
1
6
5
0
6
1
3
0
5
9
9
4
3
7
4
0
0
3
:
0
6
4
:
4
0
8
12
2
1
:
5
3
2
14
:
4
2
2
3
3
16
:
7
8
1
4
5
3
10
16
6
:
3
0
EXERCISE 9.1
Find the square roots of following
1. 4624
2. 43264
3. 99225
4. 88804
5. 488601
6. 622521
7. 63792169
8. 48818169
9. 4129024
10. 64368529
11. 36144144
12. 80946009
0
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High Speed Calculations
CHAPTER 10 Quick Cubes
Let us remember the formula
( a + b) 3 = a 3 + 3a 2 b + 3ab 2 + b 3
If observed carefully it can be seen that each term is in geometrical ratio with common
ratio b/a.
Using this premise, rules for cubing are obtained for 2 digit numbers
1. Write the cube of first digit
2. The next three terms should be in geometric ratio between second digit and first
digit of number to be cubed.
3. Now under second and third terms, double these terms and add them together.
4. Using normal rules for addition, the answer obtained is the cube.
The example would serve to illustrate the procedure completely.
123
1
2
4
4
8
1
6
12
8
1
7
2
8
nd
double 2
rd
and 3 terms
carry over 1
Numbers written in
geometric proportion,
which is 2 in this case
8
A few more examples,
523
125
50
20
8
100
40
125
150
60
8
140
6
0
8
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High Speed Calculations
273
343
8
28
98
343
56
196
8
84
294
343
19
6
8
3
27
36
48
64
72
96
27
108
144
64
39
3
0
4
Usually cubes of more than 2 digits are not really asked for in competitive examinations,
hence getting adept at this would serve all purposes.
EXERCISE 10.1
Find the cubes of following
14
23
35
37
76
38
65
57
67
47
43
81
NUMBERS CLOSE TO BASE
For numbers close to base, whether surplus or deficient, a very simple and nice way is
explained ahead.
The steps are:
1. The surplus or deficient number is doubled and added to original number. In case
of deficient it is actually adding of negative number or subtraction. This is the left
most part of answer
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2. This new surplus or deficient is multiplied by original surplus or deficient, and put
down encompassing same number of digits as that of zeroes in base as second
portion.
3. Now place cube of original surplus or deficient as last portion of answer.
Cube of 102
Base 100
Double 2 i.e. 4
surplus 2
add to 102
6 is the new surplus
Multiply 6 with 2 and put
down as 2nd portion
106/12
Cube the original surplus 2
place as 3rd portion 08 since
base 100 has 2 zeroes
106/12/08
102+4=106
Answer 1061208
Taking example of deficient, cube of 997
Base 1000
Double -3 i.e. -6
deficiency -3
add to 997
-9 is the new deficiency
Multiply -9 with -3 and put
down as 2nd portion
991/027
Cube the original surplus -3
place as 3rd portion -027 since
base 1000 has 3 zeroes
991/027/-027
Conventional calculation
027000
-027
A few more solved examples, solve them yourself also.
1043 = 112/48/64
1123 = 136/432/1728 = 1404928
9963 = 988/048/-064 = 998047936
963 = 88/48/-64 = 884736
99913 = 9973/0243/-0729 = 997302429271
997+(-6)=991
Answer 991026973
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EXERCISE 10.2
Find the cubes of following
1. 98
2. 105
3. 113
4. 995
5. 1004
6. 1012
7. 9989
8. 10007
9. 10011
10. 9992
11. 10021
12. 998
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High Speed Calculations
CHAPTER 11 Cube Roots
CUBE ROOTS UP TO 6 DIGITS
Divide the number in two parts, put a separator after the first three digits e.g. 262,144.
The left part 262 is between 216 = 63 and 343 = 73. This means 262144 lies somewhere
between 216000 = 603 and 343000 = 703. Hence the answer is surely between 60 and 70.
Something very special about cubes of single digits is that the last digit is unique. To
understand let us see the following table, the second row is cube of the first row number
respectively.
1
1
2
8
3
27
4
64
5
6
7
8
9
125 216 343 512 729
So all you have to know, rather remember very definitely is these cubes and the
associated last digit which is unique.
For 1,4,5,6 and 9 it is the same digit.
For 3,7 and 2,8 last digits are reverse of the set.
Not a tough task at all after a few minutes of practice.
Applying this to our cube 262144 in just 15 seconds the answer 64
Lets try finding cube root of 571787.
83 » as 571 is between 512 and 729 i.e. first digit of cube root is 8.
Since 7 is the last digit associated with cube of 3, therefore we have 3.
This is actually mental, calculation without need of pencil and paper.
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GENERAL CUBE ROOTS
Certain rules about cubes:
ƒ
ƒ
ƒ
ƒ
the last digit of the cube root is always distinct as explained above.
The number of digits in a cube root is same as the number of three digit
groups in the original cube.
The first digit of the cube root is from the first group in the cube.
After finding the first digit and the last digit, the work of extracting the cube
root begins.
Let F be the first digit, L the last digit and n be the number of digits in the cube root.
Remember we are dealing with exact cubes.
Let us consider a few exact cubes
Exact cube
F
L
n
39,304
3
4
2
2,299,968
1
2
3
278,445,077
6
3
3
35,578,826,569
3
9
4
The decimal expansion of any number is
a + 10b + 100c +1000d + 10000e and so on, using the various parts of the cube of this
algebraic expression, a procedure is formulated.
Let the digits of the cube be denoted by a,b,c,d,e so on from last to first respectively, i.e.
if the number is 2345 then a=5, b=4, c=3, d=2.
1. Subtract a3 from the unit's place, this eliminates the last digit.
2. Now subtract 3a2b, eliminating the last but one digit.
3. Subtract 3a2c + 3ab2, this eliminates the third last digit.
4. Now from the thousands place deduct b2 + 6abc, this is continued till required.
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All this may seem very complex, but a few examples will show how convenient it is,
Cube root of 74,618,461
Here cube root will be 3 digit.
The first digit is 4 as 43=64 and 53=125 and 74 is between
64 and 125.
Last digit, a = 1. Subtracting 1³ from last digit
74 618 461
1
3a²b = 3x1²xb, this should be ending in 6 as it is the second
last digit of the cube, thus b=2
So all the three digits are known and cube root is
74 618 46
421
Another Example
Cube root of 178,453,547
Here cube root will be 3 digit.
The first digit is 5 as 53=125 and 63=216 and 178 is
between 125 and 216.
Last digit, a = 3. Subtracting 3³ i.e. 27 from last digit
27
3a²b = 3x3²xb, this should end in 2 as it is the second last
digit of the cube, thus b = 6 as 27 x 6 = 162, ending in 2
So all the three digits are known and cube root is
178 453 547
178 453 52
563
Let us now consider a cube of 4 digit number
Cube root of 180,170,657,792
Here cube root will be 4 digit.
The first digit is 5 as 53=125 and 63=216 and 180 is
between 125 and 216.
180 170 657 792
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High Speed Calculations
Last digit, a = 8. Subtracting 8³ i.e. 512 from last digit
3a²b = 3x8²xb, this should end in 8 as 8 is the second
last digit of the cube, thus b = 4 as 192x4 = 768, ending
in 8
512
180 170 657 28
Subtracting 768 from the remaining number
3a²c+3ab² should end in 6,
7 68
180 170 49 6
3x8²xc + 3x8x4² = 192c+384, thus c=6 since this
expression will end in 6
Now all the four digits are known
5648
A few quick examples, try them yourself and compare the results,
143,877,824
First digit 5, last digit 4
4³=64
3x4²xb should end in 6, hence b=2 or 8, but
integer sum leads to 524 and not 584
143 877 824
-64
143 877 76
so cube root = 524
674,526,133
First digit 8, last digit 7
7³=343
3x7²xb should end in 9, hence b=7
So cube root = 877
674 526 133
-343
674 525 79
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High Speed Calculations
921,167,317
First digit 9, last digit 3
921 167 317
3³=27
-27
3x3²xb should end in 9, hence b=7
921 167 29
So cube root = 973
45,384,685,263
First digit 3, last digit 7
45 384 685 263
7³=343
-343
3x7²xb should end in 2, hence b=6
…….84 92
subtracting 147x6
8 82
3a²c+3ab² should end in 1
76 1
147c+756 ends 1 with c= 5
Cube root = 3567
Note: There may be cases where in step 2 for ten's digits more than one number may
suffice the requirement, this slight ambiguity can be removed by verifying the first digit
after continuing the next step. For example, 108.b ending in 2, here both 4 and 9 can be
the answers. So some intelligent guessing is a must, which is a key to quick calculations.
And most importantly use integer sum for verifying the result.
EXERCISE 11.1
Find the cube roots of following
1. 941192
2. 474552
3. 14348907
4. 40353607
5. 91733851
6. 961504803
7. 180362125
8. 480048687
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CHAPTER 12 Recurring Decimals
Let us review concept of fraction. A fraction number is one, which has a numerator and
denominator.
Numerator
Denominator
Fraction =
Some very common problems are associated with finding the decimal equivalent of
fractions. It is quite a cumbersome process if the conventional method is used. Recurring
decimals have a set of numbers after the decimal, which repeat themselves. For example
1/3 = 0.333333…… , 1/9 = 0.11111……..
FOR DENOMINATORS ENDING IN 9
There are certain observations regarding numbers ending in 9,
1. The recurring decimal will always end in 1.
2. The part of denominator except the 9 is increased by 1 to make the operator i.e.
in case of 19, the operator is 1+1=2, for 29 it is 2+1=3.
3. Put down 1 as last digit and start multiplying it by the operator from right towards
left. If the intermediate result is two digit, it is normally carried to the next
product and added.
4. When this product is equal to difference between denominator and numerator,
half the result is achieved.
5. The remaining half is found by writing down complements from 9, in simple
terms the addition of terms should be 9.
Let us start with 19,
operator is 1+1=2
Putting 1 as last digit and continually multiplying by 2 from right to left
……9 14 7 13 16 8 4 2 1
STEPS
1x2=2
2x2=4
6x2+1=13, keep 3 carry over 1
4x2+1=9
4x2=8
3x2+1=7
8x2=16, keep 6 carry over 1
7x2=14, keep 4 carry over 1
9x2=18 this is equal to 19-1, thus stop here. Half the result is found.
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947368421
052631578
complements from 9
Thus 1/19 = . 0 5 2 6 3 1 5 7 8 9 4 7 3 6 8 4 2 1 ……….(repeating this set).
Now we shall calculate 1/29
2+1=3, thus operator is 3
………19 16 15 5 21 7 12 4 11 23 27 9 3 1
This ends at 9 because 3x9+1=28 which is equal to 29-1.
Finding complements from 9, the first half is found
96551724137931
03448275862068
Thus 1/29= . 0 3 4 4 8 2 7 5 8 6 2 0 6 8 9 6 5 5 1 7 2 4 1 3 7 9 3 1 … …
FOR DENOMINATORS ENDING IN OTHER THAN 9
Now we shall involve fractions with denominators like 1, 3 or 7.
1. The last digit of the recurring decimal is such that the product of last digit of
denominator and this last digit of decimal is 9. Thus denominators ending in 7,3
and 1 will yield decimals ending with 7,3 and 9 as 7x7=49, 3x3=9 and 1x9=9.
2. Convert the fraction into one with 9 ending denominator. i.e. 1/7 can be written as
7/49.
3. This resulting denominator will serve to find the operator. 1 added to first digit, in
case of 49, 1+4=5.
4. The same method as explained above is then applied.
The simplest case of 1/7 is taken
1/7 = 7/49
Last digit will be 7, and 5 will be the operator
Starting with 7 at the right end.
28 35
7
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Now 8x5+2=42 is equal to 49-7, using complements rule
1/7= 0.142 857
Let us take another example of 1/23
1/23 = 3/69, thus 7 is the operator
39453615125127639213
Putting 3 as last digit
This ends as 7x9+3=66 = 69-3, using complements,
1/23 = 0 . 0 4 3 4 7 8 2 6 0 8 6 9 5 6 5 2 1 7 3 9 1 3 … … …
EXERCISE 12.1
Find the recurring decimals
1. 1/13
2. 2/13
3. 1/17
4. 4/7
5. 2/19
6. 1/69
7. 1/89
8. 3/23
9. 2/49
High Speed Calculations
75
CHAPTER 13 Divisibility Rule For Any Number
A very important aspect of Arithmetic in School is the rule for divisibility. There are
standard rules for numbers like 2,3,4,5,6,8,9,11 etc. But for numbers like 7,12 and so on
no rules are found in textbooks.
Schooling teaches a few basic rules which are summarized thus,
2
If the last digit divisible by two, then number is too
3
If the sum of the digits of the number is divisible by three, then number is
too
4
If the last two digits are divisible by four, then the number is too
5
If the last digit is 5 or 0, then it is divisible by 5
6
If is divisible by 2 and by 3, then number is divisible by 6
8
If the last three digits are divisible by 8, then number is too
9
If the sum of the digits of the number is divisible by nine, then it is too
10
If the last digit is 0, then number is divisible by 10
11 also has a rule which is slightly complicated. Add up all the even placed digits and
odd placed digits. Subtract these two sums and if the result is 0 or divisible by 11, then
the number is divisible.
But for other numbers like 7,13,17,19 etc., there are no straight rules. In this chapter we
will discuss the divisibility rule for any number, by any number the author means any
number. The method is adopted from the Vedic mathematics Sutra called Ekādhikena
Purveņa and sub sutra Veşţanam.
It is the method of osculation using the Veşţana, finding the Ekādhika of the divisor. We
shall not go into the details for calculating the Ekādhika of the divisor, but explain the
procedure for checking divisibility.
Let us understand what osculation means,
Suppose you were to osculate (which means to form connecting links) 21 with 5, then
multiply the last digit with 5 i.e. 5 x 1 =5, add this to previous digit 2 and thus get 7. This
is the simplest of osculations.
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A few more illustrative examples.
4321 with 7
432 + 7x1 = 439,
43+7x9=106,
10+7x6=52,
5+7x2=19,
1+7x9=64 and so on to get 34,31,10,1
NOTE: The purpose of osculation is to reach a number, which is easily identifiable
as divisible or not by the divisor.
Firstly the Ekādhika which shall be called the osculator for the numbers.
For 9,19,29,39, … they are 1,2,3,4,……respectively
For 3,13,23,33, … they are 1,4,7,10,…..respectively
For 7,17,27,37, … they are 5,12,19,26...respectively
For 1,11,21,31, … they are 1,10,19,28...respectively
To find the osculator multiply the number with the least such number that the
result has 9 as last digit. The osculator is simply 1 more than the number left after
removing its last digit 9.
For example, 13, here multiplying it by 3 we get 39, 1 more to 3 is 4, thus 4 is the
osculator.
Similarly, for 67, 67x7=469, thus 1 more than 46 i.e. 47 is the osculator.
OSCULATION BY THE Ekādhika
Let us check the divisibility of 9994 by 19. We have to osculate 9994 with 2, which is the
osculator for 19 ( 1 more than 1, first digit of 19).
The Ekādhika (osculator) for 19 is 2, so multiply the
last digit by 2 and add the product 8 to previous digit 9.
9994
999+4x2=1007
Multiply last digit of intermediate result, 7 with 2 and
add
Now multiply 4 with 2 and add with left digits 11 to get
19, thus it is proven that 9994 is divisible by 19
100+7x2=114
11+4x2=19
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Do you realize how simple it is, the only thing to remember is the osculator or the
Ekādhika.
Let us find out whether 7755 is divisible by 33 using the straight osculation method.
The Ekādhika for 33 is 10 (33x3=99 and 1 more than 9 is 10). Thus for 7755
775+50=825⇒82+50=132⇒13+20=33
A few more solved examples, try them yourself also
¾
3453 by 53, osculator is 16 {53x3=159}
345+48⇒393⇒39+48=87, thus not divisible
¾
3445 by 53, osculator is 16
344+80=424⇒42+64=106, 106=53x2, thus it is divisible
¾
191023 by 29, osculator is 3
19102+9=19111⇒1911+3=1914⇒191+12=203⇒20+9=29,divisible
¾
95591 by 17, osculator is 12 {17x9=119}
9559+12=9571⇒957+12=969⇒96+108=204=20+48=68, 68=17x4
¾
101672 by 179, osculator is 18
10167+36=10203⇒1020+54=1074⇒107+72=179, divisible
¾
5860391 by 103, osculator is 31
586039+31=586070⇒58607⇒5860+217=6077⇒607+217=824⇒82+1
24=206, divisible
EXERCISE 13.1
Check for divisibility using above method
1. 298559 by 17
2. 1937468 by 39
3. 643218 by 23
4. 2565836 by 49
5. 934321 by 21
6. 1131713 by 199
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7. 468464 by 59
8. 1772541 by 69
9. 10112124 by 33
10. 1452132 by 57
11. 777843 by 49
12. 1335264 by 21
But there is a slight problem with numbers ending in 1 and 7 the process may become
cumbersome, as the Ekādhika, OSCULATOR is a higher order number like 5,12,19,26
etc. For this another brilliant solution is the negative osculation method. Here the
principle is same except that there is subtraction instead of addition and if the end result
is a multiple of the number or zero (which is true for most cases) the divisibility is
proved.
NEGATIVE OSCULATION METHOD
If P is the positive osculator and N is the negative osculator, then the rule is divisor, D =
P+N.
For number 7, P=5, thus N=7-2=5
For number 21, P=19, thus N=21-19=2
The Negative osculators for the numbers, they are used mainly for numbers ending in 7
and 1.
For 7,17,27,37, … they are 2,5,8,11…...respectively
For 11,21,31,41, .. they are 1,2,3,4….....respectively
For 9,19,29,39,….they are 8,17,26,35.. respectively
For 3,13,23,33,…..they are 2,9,16,23….respectively
NOTE: To remember the negative osculator multiply the number to get a product
ending in 1, remove the 1 and the remaining number is the negative osculator.
The important thing to remember here is that unlike the positive osculation here osculate
by subtracting intermediate results from remaining number.
Let us consider example of testing divisibility of 165763 with 41
The negative osculator is 4.
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16576 - 4x3=16564
1656 - 4x4=1640
164 - 4x0=164
16- 4x4=0
Let us consider another example of 10171203 by 67, here multiply 67 by 3 which is 201,
hence negative osculator is 20.
1017120-20x3=1017060
101706-20x0=101706
10170-20x6=10050
1005-20x0=1005
100-20x5=0
Thus it is divisible
A few direct examples, try yourself also.
¾
358248 by 11, here negative osculator is 1 and positive osculator is 10
35824-8=35816⇒3581-6=3575⇒357-5=352⇒35-2=33
¾
1193766 by 21, here negative osculator is 2
119376-12=119364⇒11936-8=11928⇒1192-16=1176⇒117-12=105⇒
10-10=0
¾
43928 by 17, here negative osculator is 5 as 17x3=51, removing 1
4392-40=4352⇒435-10=425⇒42-25=17
¾
943093 by 37, here negative osculator is 11 as 37x3=111, removing 1
94309-33=94276⇒9427-66=9361⇒936-11=925⇒92-55=37
¾
1310598 by 51, here negative osculator is 5
131059-40=131019⇒13101-45=13056⇒1305-30=1275⇒127-25=102
⇒10-10=0
¾
387863 by 67, here negative osculator is 20 as 67x3=201
38786-60=38726⇒3872-120=3752⇒375-40=335⇒ 33-100=-67
(divisible)
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EXERCISE 13.2
Check for divisibility using above method
1. 138369 by 21
2. 204166 by 31
3. 1044885 by 41
4. 1013743 by 47
5. 1368519 by 37
6. 416789 by 37
7. 941379 by 67
8. 412357 by 81
9. 713473 by 17
10. 956182 by 17
11. 1518831 by 27
12. 426303 by 81
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High Speed Calculations
CHAPTER 14 Quick Averaging
Finding the average of a set of numbers is a very common problem in competitive
examinations as well as school level tests. It often involves long calculations and trick
division. If the weighted average is to be found out then multiplication also comes in.
This chapter would deal with simple average problems commonly encountered, using
some techniques explained earlier in this book.
The average of a set of numbers is the number, which lies in the middle of the set, or all
numbers have values centered on this average.
5,2,7,8,9,11 are 6 numbers. Now to find their average
First add them up 5+2+7+8+9+11=42 , divide this sum by 6 to get 7.
This was fairly simple but what about the numbers,
145, 136, 143, 145, 138, 137, 135, 145 set of 8 numbers.
Here it is observed that the numbers are fairly close to each other and it is easier if 140 is
chosen as the base and then the average of deviations from 140 has to be added to 140.
Average = 140 +
5-4+3+5-2-3-5+5
= 140 + 4 / 8 = 140.5
8
See how simple and easy it becomes. In most of the cases this technique would work.
Let us check a few more examples
67, 78, 81, 76, 69, 70, 81, 75, 69, 65 a set of 10 numbers.
The first number that would strike as base is 70, but if another thought is given 75 is a
better choice.
Average = 75 +
- 8 + 3 + 6 + 1 - 6 - 5 + 6 + 0 - 6 - 10
= 75 + (−19) / 10 = 73.1
10
The same method can be extended to decimal averages,
Find average of 9.9, 9.8, 10.1, 11.2, 12.0, 9.7, 10.5, 11.4
Here 10 is the most convenient base.
82
High Speed Calculations
Average = 10 +
EXERCISE 14.1
- 0.1 - 0.2 + 0.1 + 1.2 + 2.0 - 0.3 + 0.5 + 1.4
= 10 + 4.6 / 8 = 10.575
8
Find the average of the numbers using a convenient base
1. 78, 79, 84, 75, 86, 90
2. 205, 210, 215, 216, 217, 218, 220, 203
3. 5.25, 5.75, 5.45, 6.1, 5.85
4. 1768, 1890, 1798, 1759, 1803, 1802, 1750, 1765,1780,1810
Suppose you were asked to find the average weight of a class of 100 students with
following data (weight is rounded of to nearest 5)
Weight
Number of
students
40
45
50
55
60
10
11
37
23
19
A common mistake is to take the average of weight column, here comes in concept of
weighted average. Firstly find the weight of each row item by multiplying number of
students by weight.
Weight
Number of
students
Total weight
40
45
50
55
60
10
11
37
23
19
Grand total
400
495
1850
1265
1140
5150
Therefore average 5150÷100= 51.5
High Speed Calculations
83
These kinds of problems are very common in Data Interpretation section; the weighted
average needs to be calculated for table data as well as graphical data. There is no direct
shortcut, but addition and multiplication techniques explained earlier can reduce the time
taken.
84
High Speed Calculations
CHAPTER 15 Application of Techniques to Algebraic
equations
FACTORISATION OF QUADRATIC EXPRESSIONS
Let us consider the quadratic expression
2 x 2 + 2 y 2 + z 2 + 4 xy + 3 xz + 3 yz ………(1)
This is the homogeneous class quadratic. For factorization of such expressions a very
simple and easy method has been devised.
1. Put z=0 to eliminate z, retain x and y which is a simple quadratic. Factorize
this expression.
2. Now put y=0 and factorize the simple quadratic in x and z.
3. Now combine these two expressions by filling the gaps.
The ensuing steps will serve the purpose of explaining the method.
Put z=0 in (1),
2 x 2 + 2 y 2 + 4 xy ⇒ (2 x + 2 y )( x + y )
Now putting y=0 in (1)
2 x 2 + z 2 + 3 xz ⇒ (2 x + z )( x + z )
Combining these two expressions we have,
(2 x + 2 y + z )( x + y + z ) taking common coefficients.
A few more examples,
i.
put z=0 and y=0 to get expressions
2 x 2 − y 2 + 3 z 2 − xy + 5 xz − yz
(2 x + y )( x − y ) and (2 x + 3 z )( x + z ) , combining we get (2 x + y + 3 z )( x − y + z )
ii.
2 x 2 − 2 y 2 + 6 z 2 − 3 xy + 7 xz − 4 yz put z=0 and y=0 to get expressions
(2 x + y )( x − 2 y ) and (2 x + 3 z )( x + 2 z ) , combining we get (2 x + y + 3 z )( x − 2 y + z )
85
High Speed Calculations
iii.
2 x 2 − 5 y 2 − 6 z 2 + 3 xy + 4 xz − 11yz
put z=0 and y=0 to get expressions
(2 x + 5 y )( x − y ) and (2 x + 6 z )( x − z ) , combining we get (2 x + 5 y + 6 z )( x − y − z )
EXERCISE 15.1
Factorize the following quadratic expressions
1) 3 x 2 + y 2 − 2 z 2 − 4 xy − xz − yz
2) 12 x 2 − y 2 − 4 z 2 + xy + 2 xz + 3 yz
3) 2 x 2 + 2 y 2 + 6 z 2 + 4 xy − 8 xz − 8 yz
FACTORISATION OF CUBIC EXPRESSIONS
First note a very interesting and important property of factorization and factors.
The sum of coefficients in the expression(product) is equal to the product
of the sum of the coefficients of each factor.
Illustrating with the help of an example,
(x+3)(x+2)= x2+5x+6
(1+3)(1+2)=1+5+6 =12
The same holds good for all kinds of expression like cubics, biquadratics etc.
Let us consider a cubic equation,
(x+2)(x+3)(2x+1)=2x3+11x2+17x+6
(1+2)(1+3)(2+1)=2+11+17+6=36
Let us call this sum Sc (sum of coefficients).
A cubic expression is expanded thus,
( x + a )( x + b)( x + c) = x 3 + x 2 (a + b + c) + x( ab + ac + bc) + abc ……(2)
86
High Speed Calculations
This brings out some rules,
1. The coefficient of x2 is sum of the three zero power coefficients.
2. The product of these coefficients is equal to the last term of the expression.
3. If sum of coefficients of even powers is equal to sum of coefficients of odd power
then (x+1) is a factor.
4. If Sc is 0, then one of the factors is (x-1).
Let us take the example of
x3+6x2+11x+6
Here the last term is 6 whose factors are 1,1,6 or 1,2,3. But the coefficient of x2 is 6,
hence their sum should be 6. Thus 1,2,3 is the only possibility.
So factorization is (x+1) (x+2) (x+3)
Taking another example which is more complex,
Here it must be kept in mind that this is a method to speed up factorization using some
logic which could be different with different expressions, the basic equation (2) must be
memorized and the key is to manipulate the coefficients.
x3+12x2+44x+48
here a+b+c=12 and abc=48, this means a,b,c should be below 12,
Factors of 48 below 12 are 1,2,3,4,6,8 of which only 2,4,6 can satisfy both conditions of
sum and product thus factorization is (x+2)(x+4)(x+6)
This method is applicable only for expressions whose coefficient of x3 is unity.
Let us now take an example with negative coefficients,
x3+7x2+7x-15, here Sc=0, thus it is clear one of the factors is (x-1)
Now a+b+c=7 and abc=-15 since a=-1, hence the only other possibility is 3 and 5
Thus the factors are (x-1), (x+3) and (x+5).
EXERCISE 15.2
Factorize the following cubic expressions
1. x3 + 5 x 2 − 2 x − 24
2. x 3 + 9 x 2 + 20 x + 12
3. x 3 + 4 x 2 − 19 x + 14
87
High Speed Calculations
SIMULTANEOUS SIMPLE EQUATIONS
There are a fair amount of problems where simultaneous equations have to be solved.
They are usually not very complex but its does take time to solve. Here is a very useful
technique to solve equations of the type,
ax + by = m
cx + dy = n
Just remember this simple formula
x=
bn − dm
cm − an
and y =
bc − ad
bc − ad
the denominator is common and a cyclical order is used, notice for y, the lower equation
is used first i.e. cm - an .
Let us consider an example,
2 x + 3 y = −2
5x + 4 y = 2
here
3.2 − (−2.4) 6 − (−8) 14
− 2.5 − 2.2 − 10 − 4 − 14
x=
=
=
= 2 and y =
=
=
= −2
3.5 − 2.4
15 − 8
7
3.5 − 2.4
15 − 8
7
Some quick examples,
4x − 2y = 7
x + 3y = 4
here
− 2.4 − 7.3 13
7.1 − 4.4
9
x=
=
and y =
=
− 2.1 − 4.3 14
− 2.1 − 4.3 14
Taking another example,
High Speed Calculations
5 x + 2 y = 23
− 3 x + 7 y = 19
here
x=
2.19 − 23.7 123
− 23.3 − 5.19 164
=
= 3 and y =
=
=4
5.7 − (−2.3)
41
5.7 − (−2.3)
41
A few more solved examples
5x + 3 y = 2
9x + 2 y = 8
1.
here
x=
3.8 − 2.2
20
3.8 − 2.2
− 22
=
and y =
=
9.3 − (5.2) 17
9.3 − (5.2)
17
x + 9 y = −10
9 x + y = −3
2.
3.
here
− 27 − (−10) − 17
− 90 − (−27) − 63
x=
=
and y =
=
81 − 1
80
81 − 1
80
4 x + 2 y = 13
6 x + 5 y = 21
here
x=
42 − 65 23
78 − 84 6
=
=
and y =
12 − 20
8
12 − 20 8
88
89
High Speed Calculations
Solutions to Exercises in Chapters
Exercise 1.3
1.
2
8
9
7
4
3’
5’
6’
7’
8
4
3
5
4’
3
4
running
total
0
7
1
10
9
ticks
0
1
2
1
1
2
0
5
3
0
3
7
6
8
5
4’
6’
7’
9’
0
8
7’
8’
7
9’
6
Grand
total
2.
running
total
0
3
7
7
6
ticks
0
2
1
2
2
2
7
1
1
8
Grand
total
90
High Speed Calculations
3.
3
4
5
2
6
1
8’
9’
7
6’
9’
8
7
6’
8
4
5’
3’
6
7’
1
3
9’
8’
7’
running
total
0
7
6
0
7
1
ticks
0
1
2
3
2
3
2
1
1
6
2
4
2
3
4
1
5
4
6
7
9’
0
8’
8’
7’
0
0
9’
3
2
1
9’
Grand
total
4.
running
total
0
4
9
8
1
ticks
0
2
0
1
3
2
7
1
2
4
Grand
total
91
High Speed Calculations
5.
1
4
6
7
8
4
5
3
2
5’
5
8’
3’
2’
6
9’
0
0
8
7’
7’
6’
5
0
9’
running
total
0
4
1
6
8
2
ticks
0
2
2
1
1
3
2
8
4
9
2
5
Grand
total
Exercise 1.4
1.
2.
3.
4.
5.
6.
7.
361547980
376539246
546372
900876
9994563
918273645
225162
7
9
9
3
9
9
9
Exercise 3.1
1.
2.
3.
4.
Exercise 3.2
507732
346758
3413418
5405184
1.
2.
3.
4.
Exercise 3.3
1.
2.
3.
4.
173310
485465
2848815
1534420
242634
679651
3988341
2148188
Exercise 3.4
1.
2.
3.
4.
381282
1068023
6267393
3375724
Exercise 3.5
1.
2.
3.
4.
415944
1165116
6837156
3682608
92
High Speed Calculations
Exercise 3.6
Exercise 4.1
1. 311103
2. 8781564
3. 748375
4. 2887199935
5. 4811844
6. 4761988
7. 238518
8. 3994668
9. 6306104
10. 904908072
1. 156
2. 5159
3. 3618
4. 168
5. 3551
6. 7200
7. 242
8. 1044
9. 5040
10. 2670
Exercise 4.2
1.
2.
3.
4.
5.
6.
Exercise 4.3
1.
3.
5.
7.
9.
11.
8722
8536
8811
8075
9603
8742
Exercise 5.1
1.
3.
5.
7.
9.
832656
308448
89951
46420
158340
2.
4.
6.
8.
10.
471569
660516
798160
10776
89301
2.
4.
6.
8.
10.
146412
71820
864612
766878
59211
Exercise 5.2
1.
3.
5.
7.
9.
234588
111360
235900
211988
148390
10608
9483
4680
7917
525
868
2.
4.
6.
8.
10.
12.
6162
3480
9078
2160
2862
192
93
High Speed Calculations
Exercise 6.1
1.
3.
5.
7.
9.
Q=75 R=3
Q=1 R=30
Q=1 R=5569
Q=3 R=5681
Q=4 R=34579
Divide 67 by 9, Q=7, R=4
(one more way to do this)
9 67 / 8
..7 / 4
6(14 ) / 12
74 / 12
75 / 3
8998
1 / 4567
1002
.. / 1002
1 / 5569
89997
3 / 94567
10003
.. / 30009
3 / 124576
4 / 34579
2.
4.
6.
8.
10.
Q=28 R=2
Q=2 R=251
Q=6 R=145
Q=4 R=568
Q=5 R=17
Detailed solutions Exercise 6.1
9 25 / 4
..2 / 7
27 / 11
28 / 2
887
5 / 467
113
.. / 565
5 / 1032
6 / 145
86
4 / 47
14
.. / 56
4 / 103
5 / 17
89
1 / 19
11
.. / 11
997
2 / 245
003
.. / 006
1 / 30
2 / 251
8888
3 / 2345
1112
.. / 3336
3 / 5681
991
4 / 532
009
.. / 036
4 / 568
94
High Speed Calculations
Exercise 6.2
1. Q=263 R=49
3. Q=141 R=400
5. Q=135 R=449
2. Q=121 R=600
4. Q=318 R=712
6. Q=212 R=7465
Detailed solutions Exercise 6.2
89
234 / 56
11
..22
.... 5 / 5
...... /(12 )1
25 .11 /
997
121 / 237
003
.. 00 / 3
.... 0 / 06
...... / 003
...... /(10 )17
121 / 600
139 / 2174
.... 2 / − (887 × 2)
141 / 400
261 .. / 237
.
+ 2 / − (89 × 2)
263 / 49
987
314 / 578
013
.. 03 / 9
.... 0 / 13
...... / 091
887
125 / 467
113
.. 11 / 3
.... 3 / 39
898
121 / 679
102
.. 10 / 2
.... 3 / 06
...... / 408
317 / 1699
.. + 1 / − (987 )
134 / 1347
. + 1 / − (898 )
318 / 712
135 / 449
9892
210 / 4569
0108
.. 02 / 16
.... 0 / 108
...... / 0216
212 / 7465
95
High Speed Calculations
Exercise 7.1
1.
3.
5.
7.
9.
Q=20 R=3
Q=153 R=15
Q=132 R=18
Q=1136 R=33
Q=666 R=0
2. Q=21 R=0
4. Q=30 R=48
6. Q=1118 R=12
8. Q=534 R=0
10. Q=714 R=4
Detailed Solutions to Exercise 7.1
1
1
2
4
2
/
0
3
0
2
0
/
6
5
/
3
2
1
3
0
1
0
2
1
/
0
4
9
1
/
3
2
3
1
2
1
2
1
5
3
/
1
6
6
/
15
4
4
5
1
8
4
3
0
/
4
6
/
48
5
4
6
8
2
1
2
3
6
2
2
/
18
96
High Speed Calculations
6
3
5
5
0
9
1
2
6
/
4
6
3
1
1
1
8
/
8
6
3
6
/
12
7
6
7
1
3
6
9
6
1
1
3
6
/
3
8
9
8
/
33
8
3
7
3
3
2
1
5
3
4
/
4
6
3
/
0
9
2
5
3
4
4
2
1
6
6
6
/
9
2
6
/
0
10
3
8
5
3
7
3
1
6
1
4
/
4
97
High Speed Calculations
Exercise 7.2
1.
3.
5.
7.
9.
Q=142 R=4
Q=61 R=5
Q=447 R=2
Q=164 R=2
Q=468 R=8
2. Q=249 R=1
4. Q=230 R=3
6. Q=105 R=41
8. Q=126 R=34
10. Q=7912 R=13
Detailed Solutions to Exercise 7.2
1
4
2
3
1
4
1
/
2
2
1
1
4
2
/
8
2
1
/
4
2
3
3
2
4
8
2
2
4
9
/
1
4
0
/
1
3
3
2
2
8
0
6
1
/
1
4
/
5
4
8
1
4
2
2
3
0
2
3
0
/
4
9
1
/
3
5
1
1
0
4
1
4
9
0
7
/
2
98
High Speed Calculations
6
4
5
5
0
7
1
/
3
1
6
1
0
5
/
9
6
7
/
41
7
9
5
4
7
8
3
1
6
4
/
6
7
1
/
2
8
3
5
1
4
2
5
1
2
6
/
9
8
3
/
34
9
1
2
1
2
6
1
4
6
8
/
8
6
0
1
3
10
6
7
11
7
6
9
2
/
2
1
5
2
2
/
13
99
High Speed Calculations
Exercise 7.3
1.
3.
5.
7.
9.
Q=11 R=2
Q=7 R=114
Q=11 R=645
Q=21 R=125
Q=141 R=32
2. Q=11 R=9
4. Q=21 R=4216
6. Q=23 R=15
8. Q=211 R=34
10. Q=155 R=510
Detailed solutions to Exercise 7.3
1
1 1 2
-1 -2
1
2
-1
1
1
1
2
-1
/ 3
-2
-1
/ 0
4
-2
2
2
1 1 2
-1 -2
1
1
/ 4
-2
-1
/ 1
1
1
1
/ 0
9
1
2
-6
4
-4
/ 3
0
24
/ 27
0
4
+1
-16
0
(10-4+1)
7
/ 11
4
2
3
-2
/ 9
-4
/ -1
/ 4
-2
-1
3
1 6 0
-6 0
st
1
1
4
1 1 2 0 3
-1 -2 0 3
2
1
4
0
-2
2
7
-6
0
1
9
-3
6
100
High Speed Calculations
5
1 1 0 4
-1 0 -4
1
2
-1
1
1
2
7
-4
/ 7
0
-1
/ 6
8
-4
0
4
9
/ 8
-2
-6
/0
6
-2
-3
1
/1
0
0
/1
8
-6
0
2
-3
5
9
-6
-2
/ 8
7
-3
-2
/ 3
-3
4
/ 4
2
0
-1
/ 3
0
2
-4
5
6
1 2 1 1
-2 -1 -1
2
3
2
1
0
8
-3
5
7
1 0 0 3
0 0 -3
2
1
2
5
-4
8
8
1 2 3
-2 -3
2
1
1
1
5
-1
5
0
-4
9
1 1 0
-1 0
1
4
1
1
8
-2
9
-2
-12
10
1 2 2 1
-2 -2 -1
1
6
-5
/ 7
-1
-12
10
/ 4
1
5
5
/ 5
6
5
-6
10
10
5
10
1
0
Note: 4x100+10x10+10x1=400+100+10=510
101
High Speed Calculations
Exercise 7.4
1.
3.
5.
7.
9.
Q=13 R=200
Q=36 R=239
Q=11 R=378
Q=4 R=699
Q=15 R=1172
2. Q=8 R=77
4. Q=26 R=228
6. Q=3 R=667
8. Q=24 R=489
10. Q=24 R=746
Detailed solutions to Exercise 7.4
1
9 8 8
0 1 2
1
3
0
1
3
/0
1
0
/1
1
3
/2
7
/1
7
/8
4
2
3
9
4
6
10
0
0
2
8 7 8
1 2 2
7
7+1=8
0
1
14
14
14
15
800+140+15=955
955-878=77
3
8 7 9
1 2 1
3
1
3
3
4
/8
6
4
/ 18
34+2=36
8
3
3
8
4
19
7
1800+190+7=1997
1997-(879x2)=239
4
7 9 9
2 0 1
2
2
1
4
5
25+1=26
/0
0
10
/ 10
0
2
0
2
2
5
7
1027-799=228
102
High Speed Calculations
5
8 8 3
1 2 -3
1
0
1
1
/0
2
1
/3
1
9
-3
2
8
1
-3
-2
300+80-2=378
6
6 7 2
3 2 8
3 3 -2
2
/6
6
2
/ 12
2+1=3
8
6
3
-4
14
-1
1200+140-1=1339
1339-672=667
Note: the operators have been modified to ease calculations
7
9 8 2 0
0 1 8 0
0 2 -2 0
3
/9
0
9
6
7
-6
9
0
3
/ 9
15
1
9
3+1=4
10519-9820=699
8
8 1 8
1 8 2
2 -2 2
2
0
4
2
/1
-4
8
/5
4
2
4
-8
-2
1
8
9
489
9
8 8 9 3
1 1 0 7
1 1 1 -3
1
3
1
1
4
14+1=15
/4
1
4
/ 9
5
6
7
1
-3
4
4
-12
10
7
-5
9000+1000+70-5=10065
10065-8893=1172
103
High Speed Calculations
10
9 8 7
0 1 3
2
2
4
0
4
/4
2
0
/6
3
6
4
13
4
12
16
600+130+16=746
Exercise 8.1
1.
97-3/09=9409
2.
108+8/64=11664
3.
115+15/225=13225
4.
998-2/004=996004
5.
899-101/10201=798000+10201=808201 (be careful here)
6.
989-11/121=978121
7.
9987-13/0169=99740169
8.
9979-21/0441=99580441
9.
10032+32/1024=100641024
10.
1023+23/529=1046529
11.
1012+12/144=1024144
12.
99997-3/00009=9999400009
Exercise 8.2
1.
Take base 70=10x7; 68-2;66x7/4=4624
2.
Take base 200=100x2;208+8;216x2/64=43264
3.
Take base 315=100x3;315+15;330x3/225=99225
4.
Take base 300=100x3;298-2;296;296x3/04=88804
5.
Take base 800=100x8;789-11;778;778x8/121=622521
6.
Take base 700=100x7;699-1;698;698x7/01=488601
7.
Take base 8000=1000x8;7987-13;7974;7974x8/169=63792169
104
High Speed Calculations
8.
Take base 7000=1000x7;6987-13;6974;6974x7/169=48818169
9.
Take base 20000=10000x2;20032+32;20064;20064x2/01024=4012801024
10.
Take base 8000=1000x8;8023+23;8046;8046x8/529=64368529
11.
Take base 6000=1000x6;6012+12;6024;6024x6/144=36144144
12.
Take base 90000=10000x9;89997-3;89994x9/0009=8099460009
Exercise 8.3
1.
682 = .68 ⇒ 36 96 64 ⇒ 4624
2.
2082 = ..208 ⇒ 4 0 32 0 64 ⇒ 43264
3.
3152 = ..315 ⇒ 9 6 31 10 25 ⇒ 99225
4.
2982 = ..298 ⇒ 4 36
5.
6532 = ..653 ⇒ 36 60 61 30 9 ⇒ 426409
6.
7892 = ..789 ⇒ 49
7.
71412 = ...7141 ⇒ 49 14 57 22 18 8 1 ⇒ 50993881
8.
65072 = ...6507 ⇒ 36 60 25 84 70 0 49 ⇒ 42341049
9.
213132 = ….21313 ⇒ 4 4 13 10 23 12 19 6 9 ⇒ 454243969
10.
86172 = ...8617 ⇒ 64 96 52
124 85 14 49
⇒ 74252689
11.
55382 = ...5538 ⇒ 25 50 55
110 89 48 64
⇒ 30669444
12.
813472 = ….81347 ⇒ 64 16 49 70
113 144 64
⇒ 88804
112 190 144 81
⇒ 622521
129 38 58 56 49
⇒ 6617334409
Exercise 9.1
1.
4.
7.
10.
68
298
7987
8023
2.
5.
8.
11.
208
699
6987
6012
3.
6.
9.
12.
315
789
2032
8997
105
High Speed Calculations
Detailed solutions to Exercise 9.1
1
4
6
:
2
4
10
12
6
6
8
:
0
3
2
:
6
2
4
:
0
4
2
3
4
0
6
0
8
:
0
0
9
2
:
2
5
3
9
:
0
6
3
3
1
2
1
5
:
0
0
8
8
:
0
4
4
8
:
4
4
2
12
9
15
8
6
:
0
6
:
0
5
4
8
:
8
12
12
6
20
9
0
17
9
:
1
8
0
0
106
High Speed Calculations
6
6
2
:
2
5
13
14
7
:
20
2
1
15
8
8
9
:
0
0
7
9
:
2
1
7
6
3
:
14
14
7
21
26
9
8
8
1
:
20
7
:
6
9
11
0
4
0
0
6
9
8
4
8
:
12
12
6
:
20
9
8
1
24
8
:
20
7
11
:
0
0
:
4
0
0
9
4
:
1
2
0
4
2
:
1
0
9
0
3
1
2
:
:
8
2
4
1
0
0
0
0
10
6
4
:
3
0
16
8
6
3
5
4
0
2
1
4
:
0
3
:
2
1
0
9
0
0
0
4
4
11
3
6
:
0
12
6
1
0
:
4
1
2
1
0
2
:
:
0
0
0
0
0
107
High Speed Calculations
12
8
0
:
9
16
16
8
4
25
9
:
6
0
29
:
22
9
7
:
0
13
0
9
4
0
Exercise 10.1
1.
4.
7.
10.
2744
50653
274625
103823
2.
5.
8.
11.
12167
438976
185193
79507
3.
6.
9.
12.
Exercise 10.2
1.
983 = 94/12/-08 = 941192
2.
1053 = 115/75/125 = 1157625
3.
1133 = 139/507/2197 = 1442897
4.
9953 = 985/075/-125=985074875
5.
10043 = 1012/048/064=1012048064
6.
10123 = 1036/432/1728=1036433728
7.
99893 = 9967/0363/-1331=996703628669
8.
100073 = 10021/0147/0343=1002101470343
9.
100113 = 10033/0363/1331 = 1003303631331
10. 99923 = 9976/0192/-0512 = 997601919488
11. 100213 = 10063/1323/9261 = 100631329261
12. 9983 = 994/012/-008 = 994011992
42875
54872
300763
531441
0
High Speed Calculations
108
Exercise 11.1
1.
941,192 here 941 is more than 93, thus first digit is 9, and last digit is 8
as it ends in 2. Cube root is 98.
2.
474,552 here 474 is more than 73, thus first digit is 7, and last digit is 8
as it ends in 2. Cube root is 78.
3.
14,348,907
First digit c is 2, last digit a is 3, eliminating last digit 907-27=880
Now 3a2b should end in 8, so 3x 32 b = 27b ends in 8 if b=4
Thus cube root is 243
4.
40,353,607
c=3, a=3, eliminating last digit 607-27=580
Now 3a2b should end in 8, so 3x 32 b = 27b ends in 8 if b=4
Thus cube root is 343
5.
91,733,851
c=4, a=1, eliminating last digit 5 is left
Now 3a2b should end in 5, so 3 b ends in 5 if b=5
Thus cube root is 451
6.
961,504,803
c=9, a=7, eliminating last digit 803-343=460
Now 3a2b should end in 6, so 3x49 b= 147b ends in 6 if b=8
Thus cube root is 987
7.
180,362,125
c=5, a=5, eliminating last digit 125-125=000
Now 3a2b should end in 0, so 3x25 b= 75b ends in 6 if b=0,2,4,6 or 8
Here lies a confusion.
Integer sum of 180362125 is 1
Thus cube root is 5_5, where _ could be 0 or 6 as integer sum of 5053 =
13 =1 and of 5653 = 73 =343= 1.
With an intelligent guess(which is a must in quick maths) answer is 565
8.
480,048,687
c=7, a=3, eliminating last digit 687-27=660
Now 3a2b should end in 6, so 3x9 b= 27b ends in 6 if b=8
Thus cube root is 783
109
High Speed Calculations
Exercise 12.1
1.
0.076923
2.
0.153846
3.
0.0588235294117647
4.
0.571428
5.
0.105263157894736842
6.
0.0144927536231884057971
7.
0.01123595505617977528089887640449438202247191
8.
0.1304347826086956521739
9.
0.04081632653061224489755102
Exercise 13.1
1. 298559 by 17 not divisible
The ekadhika for 17 is 12
29855+108=29963⇒2996+36=3032⇒303+24=327⇒32+84=116⇒11+72=83⇒
8+36=44 is not divisible by 17
2. 1937468 by 39 not divisible
The ekadhika for 39 is 4
193746+32=193778⇒19377+32=19409⇒1940+36=1976⇒197+24=221⇒22+
4=26 is not divisible by 39
3. 643218 by 23 divisible
The ekadhika for 23 is 7
64321+56=64377⇒6437+49=6486⇒648+42=690⇒69 is divisible by 23
4. 2565836 by 49 divisible
The ekadhika for 49 is 5
256583+30=256613⇒25661+15=25676⇒2567+30=2597⇒259+35=294⇒29+
20=49 is divisible by 49
High Speed Calculations
110
5. 934321 by 21 not divisible
The ekadhika for 21 is 19
93432+19=93451⇒9345+19=9364⇒936+76=1012⇒101+38=139 is not
divisible by 21
6. 1131713 by 199 divisible
The ekadhika for 199 is 20
113171+60=113231⇒11323+20=11343⇒1134+60=1194⇒119+80=199 is
divisible by 199
7. 468464 by 59 not divisible
The ekadhika for 59 is 6
46846+24=46870⇒4687+0=4687⇒468+42=510⇒51 is not divisible by 59
8. 1772541 by 69 divisible
The ekadhika for 69 is 7
177254+7=177261⇒17726+7=17733⇒1773+21=1794⇒179+28=207⇒20+49
=69 is divisible by 69
9. 10112124 by 33 divisible
The ekadhika for 33 is 10
1011212+40=1011252⇒101125+20=101145⇒10114+50=10164⇒1016+40⇒
1056⇒105+60=165⇒16+50=66 is divisible by 33
10. 1452132 by 57 divisible
The ekadhika for 57 is 40
145213+80=145293⇒14529+120=14649⇒1464+360=1824⇒182+160=342⇒
34+80=114 is divisible by 57
High Speed Calculations
111
11. 777843 by 49 not divisible
The ekadhika for 49 is 5
77784+15=77799⇒7779+45=7824⇒782+20=802⇒80+10=90 is not divisible
by 49
12. 1335264 by 21 divisible
The ekadhika for 21 is 19
133526+76=133602⇒13360+38=13398⇒1339+152=1491⇒149+19=168 is
divisible by 21
Exercise 13.2
1. 138369 by 21 divisible
The negative osculator for 21 is 2
13836-18=13818⇒1381-16=1365⇒136-10=126⇒12-12=0 is divisible by 21
2. 204166 by 31 divisible
The negative osculator for 31 is 3
20416-18=20398⇒2039-24=2015⇒201-15=186⇒18-18=0 is divisible by 31
3. 1044885 by 41 divisible
The negative osculator for 41 is 4
104488-20=104468⇒10446-32=10414⇒1041-16=1025⇒102-20=82⇒8-8=0 is
divisible by 41
4. 1013743 by 47 divisible
The negative osculator for 47 is 14
101374-42=101332⇒10133-28=10105⇒1010-70=940⇒94 is divisible by 47
High Speed Calculations
112
5. 1368519 by 37 divisible
The negative osculator for 37 is 11
136851-99=136752⇒13675-22=13653⇒1365-33=1332⇒133-22=111⇒1111=0 is divisible by 37
6. 416789 by 37 not divisible
The negative osculator for 37 is 11
41678-99=41579⇒4157-99=4058⇒405-88=317 is not divisible by 37
7. 941379 by 67 not divisible
The negative osculator for 67 is 20
94137-180=93957⇒9395-140=9255⇒925-100=825⇒82-100=-18 is not
divisible by 67
8. 412357 by 81 not divisible
The negative osculator for 81 is 8
41235-56=41179⇒4117-72=4045⇒404-80=364⇒36-32=4 is not divisible by
81
9. 713473 by 17 divisible
The negative osculator for 17 is 5
71347-15=71332⇒7133-10=7123⇒712-15=697⇒69-35=34 is divisible by 17
10. 956182 by 17 divisible
The negative osculator for 17 is 10
95618-10=95608⇒9560-40=9520⇒952⇒95-10=85 is divisible by 17
113
High Speed Calculations
11. 1518831 by 27 divisible
The negative osculator for 27 is 8
151883-8=151875⇒15187-40=15147⇒1514-56=1458⇒145-64=81⇒8-8=0 is
divisible by 27
12. 426303 by 81 divisible
The negative osculator for 81 is 8
42630-24=42606⇒4260-48=4212⇒421-16=405⇒40-40=0 is divisible by 81
Exercise 14.1
1.
Choose base as 80, Av = 80 +(-2-1+4-5+6+10)/6 = 82
2.
Choose base as 210, Av =210 +(-5+0+5+6+7+8+10-7)/8 = 213
3.
Choose base as 5.5, Av = 5.5 +(-0.25+0.25-0.05+0.6+0.35)/5 = 5.68
4.
Choose base as 1800, Av = 1800 +(-32+90-2-41+3+2-50-35-20+10)/10=1792.5
Exercise 15.1
1. (x-y-z)(3x-y+2z)
2. (4x-y+2z)(3x+y-z)
3. (x+y-z)(2x+2y-6z)
2. (x+7)(x-2)(x-1)
3. (x+6)(x+2)(x-1)
Exercise 15.2
1. (x+4)(x+3)(x-2)